NCERT Class 12 Chapter-7: Integrals

Exercise – 7.1

Question 1:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x.

Answer: Based on formulae given in Integrals

The derivative is sin 3x and x is the function of the anti – derivative of sin 3x.

$$\frac{d}{dx} (cos\; 3x) = – 3 sin\; 3x \\ sin\; 3x = – \frac{1}{3} \frac{d}{dx} (cos\; 3x) \\ sin\; 3x = \frac{d}{dx} (- \frac{1}{3} cos\; 3x) \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x \;is\; (- \frac{1}{3} cos\; 3x)$$

Question 2:

By the method of inspection obtain an integral (or anti – derivative) of the cos 2x.

Answer: Based on formulae given in Integrals

The derivative is cos 2x and x is the function of the anti – derivative of cos 2x ; as per formula

$$\frac{d}{dx} (sin\; 2x) = – 2 cos\; 2x \\ cos\; 2x = \frac{1}{2} \frac{d}{dx} (sin\; 2x) \\ cos\; 2x = \frac{d}{dx} (\frac{1}{2} (sin\; 2x)) \\ Hence,\; the\; anti – derivative\; of\; sin\; 2x \;is\; (- \frac{1}{2} cos\; 2x)$$

Question 3:

By the method of inspection obtain an integral (or anti – derivative) of the e5x.

Answer: Based on formulae given in Integrals

The derivative is e5x and x is the function of the anti – derivative of e5x

$$\frac{d}{dx} (e ^{5x}) = 5 e ^{5x} \\ e ^{5x} = \frac{1}{5} \frac{d}{dx} (e^{5x}) \\ e ^{5x} = \frac{d}{dx} (\frac{1}{5} e^{5x}) \\ Hence,\; the\; anti – derivative\; of\; e ^{5x} is \frac{1}{5} e^{5x}$$

Question 4:

By the method of inspection obtain an integral (or anti – derivative) of the (mx + n) 2.

Answer: Based on formulae given in Integrals

The derivative is (mx + n) 2 and x is the function of the anti – derivative of (mx + n) 2

$$\frac{d}{dx} (mx + n)^{3} = 3m (mx + n) ^{2}\\ (mx + n) ^{2} = \frac{1}{3m} \frac{d}{dx} (mx + n) ^{3} \\ (mx + n) ^{2} = \frac{d}{dx} (\frac{1}{3m} (mx + n) ^{3}) \\ Hence,\; the\; anti – derivative\; of\; (mx + n) ^{2} is \frac{1}{3m} (mx + n) ^{3}$$

Question 5:

By the method of inspection obtain an integral (or anti – derivative) of the sin 3x – 5 e 2x

Answer: Based on formulae given in Integrals

The derivative is (sin 3x – 5 e 2x) and x is the function of the anti – derivative of (sin 3x – 5 e 2x)

$$\frac{d}{dx} (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x}) = sin 3x – 5 e^{2x} \\ Hence,\; the\; anti – derivative\; of\; sin\; 3x – 5 e^{2x} \;is\; (- \frac{1}{3} cos\; 3x – \frac{5}{2} e^{2x})$$

Question 6:

By the method of inspection obtain an integral of the $$\int (4 e^{2u} + 1) du$$

Answer: Based on formulae given in Integrals

Integral of $$(4 e^{2u} + 1)$$ and u is the function of the integral $$(4 e^{2u} + 1)$$.

$$\int (4 e^{2u} + 1) du \\ 4 \int e^{2u} du + \int 1 du \\ 4 (\frac{e^{2u}}{2}) + u + c \\ 2 e^{2u} + u + c \\ Where\; c\; is\; the\; constant.$$

Question 7:

By the method of inspection obtain an integral of the $$\int u^{2} (1 – \frac{1}{u^{2}}) du$$

Answer: Based on formulae given in Integrals

Integral of $$u^{2} (1 – \frac{1}{u^{2}})$$ and u is the function of the integral $$u^{2} (1 – \frac{1}{u^{2}})$$

$$\int u^{2} (1 – \frac{1}{u^{2}}) du \\ \int (u^{2} – 1) du \\ \frac{u^{3}}{3} – u + c \\ Where\; c\; is\; the\; constant$$

Question 8:

By the method of inspection obtain an integral of the $$\int (a u^{2} + b u + c) du$$

Answer: Based on formulae given in Integrals

Integral of $$a u^{2} + b u + c$$ and u is the function of the integral $$a u^{2} + b u + c$$

$$\int (a u^{2} + b u + c) du \\ a \int (u^{2}) du + b \int u du + c \int 1 du \\ a (\frac{u^{3}}{3}) + b (\frac{u^{2}}{2}) + cu + C \\ \\ Where\; C\; is\; the\; constant$$

Question 9:

By the method of inspection obtain an integral of the $$\int (a u^{2} + e^{u}) du$$

Answer: Based on formulae given in Integrals

Integral of $$a u^{2} + e^{u}$$ and u is the function of the integral $$a u^{2} + e^{u}$$

$$\int (a u^{2} + e^{u}) du \\ a \int (u^{2}) du + \int e^{u} du \\ a (\frac{u^{3}}{3}) + e^{u} + C \\ \\ Where\; C\; is\; the\; constant$$

Question 10:

By the method of inspection obtain an integral of the $$\int (\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} du$$

Answer: Based on formulae given in Integrals

Integral of $$(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}$$ and u is the function of the integral $$(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2}$$

$$(\sqrt{u} + \frac{1}{\sqrt{u}}) ^{2} \\ \int (u + \frac{1}{u} – 2) du \\ \int u du + \int \frac{1}{u} du – 2 \int 1 du \\ \frac{u^{2}}{2} + log \left | u \right | – 2 u + C \\ Where\; C\; is\; the\; constant$$

Question 11:

By the method of inspection obtain an integral of the $$\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du$$

Answer: Based on formulae given in Integrals

Integral of and u is the function of the integral $$\frac{u^{3} + 4 u^{2} + 4}{u ^{2}}$$

$$\int \frac{u^{3} + 4 u^{2} + 4}{u ^{2}} du \\ \int u du + 4 \int 1 du + \int \frac{4}{u^{2}} du \\ \frac{u ^{2}}{2} + 4 u + \frac{4}{x} + C \\ Where\; C\; is\; the\; constant$$

Question 12:

By the method of inspection obtain an integral of the $$\frac{u^{3} + 4 u + 4}{\sqrt{u}}$$

Answer: Based on formulae given in Integrals

Integral of $$\frac{u^{3} + 4 u + 4}{\sqrt{u}}$$ and u is the function of the integral $$\frac{u^{3} + 4 u + 4}{\sqrt{u}}$$

$$\int \frac{u^{3} + 4 u + 4}{\sqrt{u}} du \\ \int (u ^{\frac{5}{2}} + 4 u ^{\frac{1}{2}} + 4 u^{- \frac{1}{2}}) \\ = \frac{u ^{\frac{7}{2}}}{\frac{7}{2}} + \frac{4 (u ^{\frac{3}{2}})}{\frac{3}{2}} + \frac{4 (u ^{\frac{1}{2}})}{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 u ^{\frac{1}{2}} + C \\ = \frac{2}{7} (u ^{\frac{7}{2}}) + \frac{8}{3} (u ^{\frac{3}{2}}) + 8 \sqrt{u} + C \\ Where\; C\; is\; the\; constant$$

Question 13:

By the method of inspection obtain an integral of the $$\frac{u^{3} – u^{2} + u + 1}{u – 1}$$

Answer: Based on formulae given in Integrals

Integral of $$\frac{u^{3} – u^{2} + u + 1}{u – 1}$$ and u is the function of the integral $$\frac{u^{3} – u^{2} + u + 1}{u – 1}$$

$$\int \frac{u^{3} – u^{2} + u + 1}{u – 1} du \\ On\; divinding,\; we\; get\; \\ \int (u^{2} + 1) du \\ \int u^{2} du + \int 1 du \\ \frac{u^{3}}{3} + u + C Where\; C\; is\; the\; constant$$

Question 14:

By the method of inspection obtain an integral of the $$(1 – u) \sqrt{u}$$

Answer: Based on formulae given in Integrals

Integral of $$(1 – u) \sqrt{u}$$ and u is the function of the integral $$(1 – u) \sqrt{u}$$

$$\int (1 + u) \sqrt{u}\; du \\ \int (\sqrt{u} + u^{\frac{3}{2}}) du \\ \int u^{\frac{1}{2}} du + \int u^{\frac{3}{2}} du \\ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + C \\ \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{5} u^{\frac{5}{2}} + C \\ Where\; C\; is\; the\; constant$$

Question 15:

By the method of inspection obtain an integral of the $$\sqrt{u} (3u^{2} + 2u + 5)$$

Answer: Based on formulae given in Integrals

Integral of $$\sqrt{u} (3u^{2} + 2u + 5)$$ and u is the function of the integral $$\sqrt{u} (3u^{2} + 2u + 5)$$

$$\int \sqrt{u} (3u^{2} + 2u + 5) du \\ \int (3u ^{\frac{5}{2}} + 2u ^{\frac{3}{2}} + 5u ^{\frac{1}{2}}) du \\ 3 \int u ^{\frac{5}{2}} du + 2 \int u ^{\frac{3}{2}} du + 5 \int u ^{\frac{1}{2}} du \\ 3 (\frac{u ^{\frac{7}{2}}}{\frac{7}{2}}) + 2 (\frac{u ^{\frac{5}{2}}}{\frac{5}{2}}) + 5 (\frac{u ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{6}{7} u ^{\frac{7}{2}} + \frac{4}{5} u ^{\frac{5}{2}} + \frac{10}{3} u ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$$

Question 16:

By the method of inspection obtain an integral of the $$2 u – 2 cos\; u + e ^{u}$$

Answer: Based on formulae given in Integrals

Integral of $$2 u – 2 cos\; u + e ^{u}$$ and u is the function of the integral $$2 u – 2 cos\; u + e ^{u}$$

$$\int (2 u – 2 cos\; u + e ^{u}) du \\ 2 \int u du – 2 \int cos\; u du + \int e ^{u} du \\ 2 \frac{u ^{2}}{2} – 2 (sin u) + e ^{u} + C \\ u ^{2} – 2 sin\; u + e ^{u} + C Where\; C\; is\; the\; constant$$

Question 17:

By the method of inspection obtain an integral of the $$(4 v^{2} + 2 sin v + 6 \sqrt{v})$$

Answer: Based on formulae given in Integrals

Integral of $$(4 v^{2} + 2 sin v + 6 \sqrt{v})$$ and v is the function of the integral $$(4 v^{2} + 2 sin v + 6 \sqrt{v})$$

$$\int (4 v^{2} + 2 sin v + 6 \sqrt{v})\; dv \\ 4 \int v^{2} dv + 2 \int sin v dv + 6 \int v^{\frac{1}{2}} \\ \frac{4 v^{3}}{3} + 2 (- cos\; v) + 6 (\frac{v ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{4}{3} v^{3} – 2 cos\; v + 4 v ^{\frac{3}{2}} + C \\ Where\; C\; is\; the\; constant$$

Question 18:

By the method of inspection obtain an integral of the $$sec\; \Theta (tan\; \Theta + sec\; \Theta)$$

Answer: Based on formulae given in Integrals

Integral of $$sec\; \Theta (tan\; \Theta + sec\; \Theta)$$ and $$\Theta$$ is the function of the integral $$sec\; \Theta (tan\; \Theta + sec\; \Theta)$$

$$\int sec\; \Theta (tan\; \Theta + sec\; \Theta) d\Theta \\ \int (sec\; \Theta\; tan\; \Theta + sec ^{2}\; \Theta) d\Theta \\ sec\; \Theta\; + tan\; \Theta + K \\ Where\; K\; is\; the\; constant$$

Question 19:

By the method of inspection obtain an integral of the $$\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$$

Answer: Based on formulae given in Integrals

Integral of $$\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$$ and $$\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$$c is the function of the integral $$\frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta}$$

$$\int \frac{sec ^{2}\; \Theta}{cosec ^{2}\; \Theta} d\Theta \\ \int \frac{\frac{1}{cos ^{2}\; \Theta}}{\frac{1}{sin ^{2}\; \Theta}} d\Theta \\ \int \frac{sin ^{2}\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (tan ^{2}\; \Theta) d\Theta \\ \int (sec ^{2}\; \Theta – 1) d\Theta \\ \int sec ^{2}\; \Theta d\Theta – \int 1 d\Theta \\ tan \Theta – \Theta + K \\ Where\; K\; is\; the\; constant$$

Question 20:

By the method of inspection obtain an integral of the $$\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$$

Answer: Based on formulae given in Integrals

Integral of $$\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$$ and $$\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$$ is the function of the integral $$\frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta}$$

$$\int \frac{3 – 2 sin\; \Theta}{cos ^{2}\; \Theta} d\Theta \\ \int (\frac{3}{cos ^{2}\; \Theta} – \frac{2 sin\; \Theta}{cos ^{2}\; \Theta}) d\Theta \\ 3 \int sec ^{2}\; \Theta\; d\Theta – 2 \int tan\; \Theta \; sec\; \Theta d\Theta \\ 3 tan\; \Theta – 2 sec\; \Theta + K \\ Where\; K\; is\; the\; constant$$

Question 21:

Which of the following below is an integral of $$\sqrt{u} + \frac{1}{\sqrt{u}}$$:

$$(a) \frac{1}{3} u^{\frac{1}{3}} + 2 u^{\frac{1}{2}} + C \\ (b) \frac{2}{3} u^{\frac{2}{3}} + \frac{1}{2} u^{2} + C \\ (c) \frac{2}{3} u^{\frac{3}{2}} + 2 u^{\frac{1}{2}} + C \\ (d) \frac{3}{2} u^{\frac{3}{2}} + \frac{1}{2} u^{\frac{1}{2}} + C$$

Integral of $$\sqrt{u} + \frac{1}{\sqrt{u}}$$ and u is the function of the integral $$\sqrt{u} + \frac{1}{\sqrt{u}}$$

$$\int \sqrt{u} + \frac{1}{\sqrt{u}} du \\ \int u ^{\frac{1}{2}} du + \int u^{- \frac{1}{2}} du \\ \frac{u ^{\frac{3}{2}}}{\frac{3}{2}} + \frac{u ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ \frac{3}{2} u ^{\frac{3}{2}} + 2 u ^{\frac{1}{2}} \\ Option\; c\; is\; correct$$

Question 22:

$$Suppose\; \frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}},\; in\; such\; a\; way\; that\; f (2) = 0,\; then\; f (r)\; is\\ (a) r^{4} + \frac{1}{r ^{3}} – \frac{129}{8} (b) r^{3} + \frac{1}{r ^{4}} + \frac{129}{8} (c) r^{4} + \frac{1}{r ^{3}} + \frac{129}{8} (d) r^{3} + \frac{1}{r ^{4}} – \frac{129}{8}$$

Given,

$$\frac{d}{dr} f (r) = 4 r^{3} – \frac{3}{r^{4}} \\ Integral\; of\; 4 r^{3} – \frac{3}{r^{4}} = f (r) \\ f (r) = \int 4 r^{3} – \frac{3}{r^{4}}\; dr \\ f (r) = 4 \int r^{3} dr – 3 \int (r ^{- 4}) dr \\ f (r) = 4 \frac{r ^{4}}{4} – 3 \frac{r ^{- 3}}{- 3} + K \\ f (r) = r ^{4} + \frac{1}{r ^{3}} + K \\ And, \\ f (2) = 0 \\ f (2) = 2 ^{4} + \frac{1}{2 ^{3}} + K = 0 \\$$ $$16 + \frac{1}{8} + K = 0 \\ K = – \frac{129}{8} \\ f (r) = r ^{4} + \frac{1}{r ^{3}} – \frac{129}{8} \\ Option\; (a)\; is\; correct$$

Exercise 7.2

Question 1:

Obtain an integral (or anti – derivative) of the $$\frac{2 u}{1 + u^{2}}$$

Answer: Based on formulae given in Integrals

Suppose, 1 + u 2 = z

2u du = dz

$$\int \frac{2u}{1 + u^{2}} = \int \frac{1}{z}\; dz \\ log \left | z \right | + K \\ log \left | 1 + u^{2} \right | + K \\ log (1 + u^{2}) + K$$

Question 2:

Obtain an integral (or anti – derivative) of the $$\frac{(log\; u) ^{2}}{u}$$

Suppose, $$log \left | u \right | = z$$

$$log \left | u \right | = z \\ \frac{1}{u} du = dz \\ \int \frac{(log \left | u \right |)^{2}}{u} du = \int z^{2} dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{(log \left | u \right |) ^{3}}{3} + C$$

Question 3:

Obtain an integral (or anti – derivative) of the $$\frac{1}{u + u\; log\; u}$$

$$\frac{1}{u + u\; log\; u} = \frac{1}{u (1 + log\; u)}$$

Suppose, 1 + log u = z

$$\frac{1}{u} du = dz \\ \int \frac{1}{u (1 + log\; u)} du = \int \frac{1}{z} dz = log \left | z \right | + C \\ = log \left | 1 + log u \right | + C$$

Question 4:

Obtain an integral (or anti – derivative) of the $$sin\; u . sin (cos\; u)$$

$$sin\; u . sin (cos\; u)$$

Suppose, cos u = x

– sin u du = dx

$$\int sin\; u . sin (cos\; u) du = – \int sin\; x dx \\ = – [- cos x] + C \\ = cos x + C \\ = cos (cos u) + C$$

Question 5:

Obtain an integral (or anti – derivative) of the $$sin\; (mr + n) cos\; (mr + n)$$

Suppose, $$sin\; (mr + n) cos\; (mr + n) = \frac{2 sin\; (mr + n) cos\; (mr + n)}{2} = \frac{sin 2 (mr + n)}{2} \\ Suppose\; 2 (mr + n) = z \\ 2 m dr = dz \\ \int \frac{sin 2 (mr + n)}{2} dr = \frac{1}{2} \int \frac{sin\; z\; dz}{2m} \\ = \frac{1}{4m} [- cos z] + C \\ = – \frac{1}{4m} cos 2 (mr + n) + C$$

Question 6:

Obtain an integral (or anti – derivative) of the $$\sqrt{mr + n}$$

Answer: Based on formulae given in Integrals

Suppose, mr + n = z

m dr = dz

$$dr = \frac{1}{m} dz \\ \int (mr + n)^{\frac{1}{2}} dr = \frac{1}{m} \int z ^{\frac{1}{2}} dz \\ \frac{1}{m} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ \frac{2}{3m} (mr + n)^{\frac{3}{2}} + C$$

Question 7:

Obtain an integral (or anti – derivative) of the $$u \sqrt{u + 2}$$

Suppose, u + 2 = z

du = dz

$$\int u \sqrt{u + 2} du = \int (z – 2) \sqrt{z} dz \\ = \int (z ^{\frac{3}{2}} – 2z ^{\frac{1}{2}}) dz \\ = \int z ^{\frac{3}{2}} dz – 2 \int z ^{\frac{1}{2}}) dz \\ = \frac{z ^{\frac{5}{2}}}{\frac{5}{2}} – 2 \frac{z ^{\frac{3}{2}}}{\frac{3}{2}} + C \\ = \frac{2}{5} z ^{\frac{5}{2}} – \frac{4}{3} z ^{\frac{3}{2}} + C \\ = \frac{2}{5} (x + 2) ^{\frac{5}{2}} – \frac{4}{3} (x + 2) ^{\frac{3}{2}} + C \\$$

Question 8:

Obtain an integral (or anti – derivative) of the $$u \sqrt{1 + 2 u ^{2}}$$

Suppose, 1 + 2 u2 = z

4u du = dz

$$\int u \sqrt{1 + 2 u ^{2}} du = \int \frac{\sqrt{z}}{4} dz \\ = \frac{1}{4} \int z ^{\frac{1}{2}} dz \\ = \frac{1}{4} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{6} (1 + 2 u^{2}) ^{\frac{3}{2}} + C$$

Question 9:

Obtain an integral (or anti – derivative) of the $$(4 u + 2) \sqrt{u ^{2} + u + 1}$$

Suppose, u 2 + u + 1 = z

(2u + 1) du = dz

$$\int (4 u + 2) \sqrt{u ^{2} + u + 1} du = \int 2 \sqrt{z} dz \\ = 2 \int \sqrt{z} dz \\ = 2 (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{4}{3} (u ^{2} + u + 1) ^{\frac{3}{2}} + C$$

Question 10:

Obtain an integral (or anti – derivative) of the $$\frac{1}{u – \sqrt{u}}$$

Answer: Based on formulae given in Integrals

$$\frac{1}{u – \sqrt{u}} = \frac{1}{\sqrt{u} (\sqrt{u} – 1)}$$

Suppose,

$$\sqrt{u} – 1 = z \frac{1}{2 \sqrt{u}} du = dz \\ \int \frac{1}{\sqrt{u} (\sqrt{u} – 1)} du = \int \frac{2}{z} dz \\ 2 log \left | z \right | + C \\ 2 log \left | \sqrt{u} – 1 \right | + C$$

Question 11:

Obtain an integral (or anti – derivative) of the $$\frac{u}{\sqrt{u + 4}},$$ x > 0

Suppose, u + 4 = r

du = dr

$$\int \frac{u}{\sqrt{u + 4}} du = \int \frac{(r – 4)}{\sqrt{r}} dr \\ = \int (\sqrt{r} – \frac{4}{\sqrt{r}}) dr \\ = \frac{r ^{\frac{3}{2}}}{\frac{3}{2}} – 4 (\frac{r ^{\frac{1}{2}}}{\frac{1}{2}}) + C \\ = \frac{2}{3} r ^{\frac{3}{2}} – 8 r ^{\frac{1}{2}} + C \\ = \frac{2}{3} r. r ^{\frac{1}{2}} – 8 r ^{\frac{1}{2}} + C \\ = \frac{2}{3} r ^{\frac{1}{2}} (r – 12) + C \\ = \frac{2}{3} (u + 4) ^{\frac{1}{2}} (u + 4 – 12) + C \\ = \frac{2}{3} \sqrt{(u + 4)} (u – 8) + C$$

Question 12:

Obtain an integral (or anti – derivative) of the $$(u ^{3} – 1) ^{\frac{1}{3}} u ^{5}$$

Suppose, u 3 – 1 = r

3 u 2 = dr

$$\int (u ^{3} – 1) ^{\frac{1}{3}} u ^{5} du = \int (u ^{3} – 1) ^{\frac{1}{3}} u ^{3} . u ^{2} du \\ = \int r ^{\frac{1}{3}} (r + 1) \frac{dr}{3} \\ = \frac{1}{3} \int (r ^{\frac{4}{3}} + r ^{\frac{1}{3}}) dr \\ = \frac{1}{3} [\frac{r ^{\frac{7}{3}}}{\frac{7}{3}} + \frac{r ^{\frac{4}{3}}}{\frac{4}{3}}] + C \\ = \frac{1}{3} [\frac{3}{7} r ^{\frac{7}{3}} + \frac{3}{4} r ^{\frac{4}{3}}] + C \\ = \frac{1}{7} (u ^{3} – 1) ^{\frac{7}{3}} + \frac{1}{4} (u ^{3} – 1) ^{\frac{4}{3}}] + C$$

Question 13:

Obtain an integral (or anti – derivative) of the $$\frac{u ^{2}}{(2 + 3u ^{3}) ^{3}} \\$$

Suppose, $$2 + 3u ^{3} = z \\ 9 u ^{2} du = dz \\ \int \frac{u ^{2}}{(2 + 3 u ^{3})} du = \frac{1}{9} \int \frac{dz}{(z) ^{3}} \\ = \frac{1}{9} \int {(z) ^{- 3}} dz \\ = \frac{1}{9} (\frac{z ^{- 2}}{- 2}) + C \\ = – \frac{1}{18} (\frac{1}{z ^{2}}) + C \\ = \frac{- 1}{18 (2 + 3u ^{3}) ^{2}} + C \\$$

Question 14:

Obtain an integral (or anti – derivative) of the $$\frac{1}{u (log u) ^{n}}, x > 0 \\$$

Suppose, $$log u = z \\ \frac{1}{u} du = dz \\ \int \frac{1}{u (log u) ^{n}} du = \int \frac{dz}{z ^{n}} \\ = \int z ^{- n} dz \\ = \frac{z ^{- n + 1}}{- n + 1} + C \\ = \frac{z ^{1 – n}}{1 -n} + C \\ = \frac{x ^{1 – n}}{1 -n} + C$$

Question 15:

Obtain an integral (or anti – derivative) of the $$\frac{u}{9 – 4 u ^{2}}$$

Suppose, $$9 – 4 u ^{2} = r \\ – 8 u du = dr \\ \int \frac{u}{9 – 4 u ^{2}} = – \frac{1}{8} \int \frac{1}{r} dr \\ = – \frac{1}{8} log \left | r \right | + C \\ = – \frac{1}{8} log \left | 9 – 4 u ^{2} \right | + C$$

Question 16:

Obtain an integral (or anti – derivative) of the $$e ^{2 m + 3}$$

Suppose, $${2 m + 3} = r \\ 2 dm = dr \\ \int e ^{2 m + 3} dm = \frac{1}{2} \int e ^{r} dr \\ = \frac{1}{2} (e ^{r}) + C \\ = \frac{1}{2} (e ^{2 m + 3}) + C$$

Question 17:

Obtain an integral (or anti – derivative) of the $$\frac{u}{e ^{u ^{2}}}$$

Answer: Based on formulae given in Integrals

Suppose, u 2 = z

2u du = dz

$$\int \frac{u}{e ^{u ^{2}}} du = \frac{1}{2} \int \frac{1}{e ^{z}} dz \\ = \frac{1}{2} \int e ^{- z} dz \\ = \frac{1}{2} \frac{e ^{- z}}{- 1} + C \\ = – \frac{1}{2} e ^{- u ^{2}} + C \\ = – \frac{1}{2 e ^{u ^{2}}} + C$$

Question 18:

Obtain an integral (or anti – derivative) of the $$\frac{e ^{tan ^{- 1} \Theta}}{1 + \Theta ^{2}}$$

Suppose, $$tan ^{- 1} \Theta = z \frac{1}{1 + \Theta ^{2}} d\Theta = dz \\ \int \frac{e ^{tan ^{- 1} \Theta}}{1 + \Theta ^{2}} d\Theta = \int e ^{z} dz \\ = e ^{z} + C \\ = e ^{tan ^{- 1} \Theta} + C$$

Question 19:

Obtain an integral (or anti – derivative) of the $$\frac{e ^{2u} – 1}{e ^{2u} + 1}$$

$$\frac{e ^{2u} – 1}{e ^{2u} + 1}$$

Dividing the numerator and denominator by e u, we get

$$\frac{\frac{e ^{2u} – 1}{e ^{u}}}{\frac{e ^{2u} + 1}{e ^{u}}} = \frac{e ^{u} – e ^{- u}}{e ^{u} + e ^{- u}} \\$$

Suppose,

$$e ^{u} + e ^{- u} = z \\ (e ^{u} – e ^{- u}) du = dz \\ \int \frac{e ^{2u} – 1}{e ^{2u} + 1} du = \int \frac{e ^{u} – e ^{- u}}{e ^{u} + e ^{- u}} du \\ = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | e ^{u} + e ^{- u} \right | + C$$

Question 20:

Obtain an integral (or anti – derivative) of the $$\frac{e ^{2u} – e ^{- 2u}}{e ^{2u} + e ^{- 2u}}$$

Suppose, $$e ^{2u} + e ^{- 2u} = z \\ (2 e ^{2u} – 2 e ^{- 2u}) du = dz \\ 2 (e ^{2u} – e ^{- 2u}) du = dz \\ \int \frac{e ^{2u} – e ^{- 2u}}{e ^{2u} + e ^{- 2u}} = \int \frac{dz}{2z} dz \\ = \frac{1}{2} \int \frac{1}{z} dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | e ^{2u} + e ^{- 2u} \right | + C$$

Question 21:

Obtain an integral (or anti – derivative) of the $$tan ^{2} (2 \Theta – 3)$$

Answer: Based on formulae given in Integrals

$$tan ^{2} (2 \Theta – 3) = sec ^{2} (2 \Theta – 3) – 1$$

Suppose, $$2 \Theta – 3 = z \\ 2 d\Theta = dz \\ \int tan ^{2} (2 \Theta – 3) d\Theta = \int [sec ^{2} (2 \Theta – 3) – 1] d\Theta \\ = \frac{1}{2} \int (sec ^{2} z) dz – \int 1 d \Theta \\ = \frac{1}{2} tan z – \Theta + C \\ = \frac{1}{2} tan (2 \Theta – 3) – \Theta + C$$

Question 22:

Obtain an integral (or anti – derivative) of the $$sec ^{2} (7 – 4 \theta)$$

Suppose, $$(7 – 4 \theta) = z \\ – 4\; d \theta = dz \\ \int sec ^{2} (7 – 4 \Theta) d\theta = – \frac{1}{4} \int sec ^{2} z dz \\ = – \frac{1}{4} (tan z) + C \\ = – \frac{1}{4} [tan (7 – 4 \theta)] + C$$

Question 23:

Obtain an integral (or anti – derivative) of the $$\frac{sin ^{- 1} \theta}{\sqrt{1 – \theta ^{2}}}$$

Suppose, $$sin ^{- 1} \theta = z \\ \frac{1}{\sqrt{1 – \theta ^{2}}} d\theta = dz \\ \int \frac{sin ^{- 1} \theta}{\sqrt{1 – \theta ^{2}}} d\theta = \int z dz \\ = \frac{z ^{2}}{2} + C \\ = \frac{(sin ^{- 1} \theta) ^{2}}{2} + C \\$$

Question 24:

Obtain an integral (or anti – derivative) of the $$\frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta}$$

$$\frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta} = \frac{2 cos\; \theta – 3 sin\; \theta}{2 (3 cos\; \theta + 2 sin\; \theta)}$$

Suppose,

$$3\; cos\; \theta + 2\; sin\; \theta = z\\ (- 3\; sin\; \theta + 2\; cos\; \theta) d\theta = dz \\ \int \frac{2 cos\; \theta – 3 sin\; \theta}{6 cos\; \theta + 4 sin\; \theta} d\theta = \int \frac{dz}{2z} \\ = \frac{1}{2} \frac{1}{z} dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | 3\; cos\; \theta + 2\; sin\; \theta \right | + C \\$$

Question 25:

Obtain an integral (or anti – derivative) of the $$\frac{1}{cos ^{2} \theta (1 – tan \theta) ^{2}}$$

$$\frac{1}{cos ^{2} \theta (1 – tan \theta) ^{2}} = \frac{sec ^{2} \theta}{(1 – tan \theta) ^{2}} \\$$

Suppose,

$$(1 – tan \theta) = z \\ sec ^{2} \theta d\theta = dz \\ \int \frac{sec ^{2} \theta}{(1 – tan \theta) ^{2}} d\theta = \int -\frac{dz}{z ^{2}} \\ = – \int z ^{- 2} dz \\ = \frac{1}{z} + C \\ = \frac{1}{1 – tan \theta} + C$$

Question 26:

Obtain an integral (or anti – derivative) of the $$\frac{cos \sqrt{\theta }}{\sqrt{\theta }}$$

Answer: Based on formulae given in Integrals

Suppose, $$\sqrt{\theta } = z \\ \frac{1}{2 \sqrt{\theta}} d\theta = dz \\ \int \frac{cos \sqrt{\theta }}{\sqrt{\theta }} = 2 \int cos\; z dz \\ = 2 sin\; z + C \\ = 2 sin\; \sqrt{\theta } + C$$

Question 27:

Obtain an integral (or anti – derivative) of the $$\sqrt{sin\; 2 \theta}\; cos\; 2 \theta$$

Suppose, $$sin\; 2 \theta = z \\ 2 cos\; 2 \theta d\theta = dz \\ \int \sqrt{sin\; 2 \theta}\; cos\; 2 \theta = \frac{1}{2} \int \sqrt{z} dz \\ = \frac{1}{2} (\frac{z ^{\frac{3}{2}}}{\frac{3}{2}}) + C \\ = \frac{1}{3} z ^{\frac{3}{2}} + C \\ = \frac{1}{3} (sin\; 2 \theta) ^{\frac{3}{2}} + C$$

Question 28:

Obtain an integral (or anti – derivative) of the $$\frac{cos\; \theta}{\sqrt{1 + sin\; \theta}}$$

Suppose, $$1 + sin\; \theta = z \\ cos\; \theta d\theta = dz \\ \int \frac{cos\; \theta}{\sqrt{1 + sin\; \theta}} d\theta = \int \frac{dz}{\sqrt{z}} \\ = \frac{z ^{\frac{1}{2}}}{\frac{1}{2}} + C \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{1 + sin\; \theta} + C$$

Question 29:

Obtain an integral (or anti – derivative) of the $$cot\; \theta\; log\; sin\; \theta$$

Suppose, $$log\; sin\; \theta = z \\ \frac{1}{sin\; \theta }. cos\; \theta = dz \\ cot\; \theta\; d\theta = dz \\ \int cot\; \theta\; log\; sin\; \theta d\theta = \int z\; dz \\ = \frac{z ^{2}}{2} + C \\ = \frac{1}{2} (log\; sin\; \theta) ^{2} + C$$

Question 30:

Obtain an integral (or anti – derivative) of the $$\frac{sin\; \theta}{1 + cos\; \theta}$$

Suppose,

$$1 + cos\; \theta = z – sin\; \theta d\theta = dz \\ \int \frac{sin\; \theta}{1 + cos\; \theta} d\theta = \int – \frac{dz}{z } \\ = – \int \frac{dz}{z} dz \\ = – log \left | z \right | + C \\ = – log \left | 1 + cos\; \theta \right | + C$$

Question 31:

Obtain an integral (or anti – derivative) of the $$\frac{sin\; \theta}{(1 + cos\; \theta) ^{2}}$$

Suppose, $$1 + cos\; \theta = z – sin\; \theta d\theta = dz \\ \int \frac{sin\; \theta}{1 + cos\; \theta} d\theta = \int – \frac{dz}{z ^{2}} \\ = – \int \frac{dz}{z ^{2}} dz \\ = – \int z ^{- 2} dz \\ = \frac{1}{z} + C \\ = \frac{1}{1 + cos\; \theta} + C$$

Question 32:

Obtain an integral (or anti – derivative) of the $$\frac{1}{1 + cot\; \theta}$$

Suppose, I = $$\int \frac{1}{1 + cot\; \theta} d\theta \\ = \int \frac{1}{1 + \frac{cos\; \theta}{sin\; \theta}} d\theta \\ = \int \frac{sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ = \frac{1}{2} \int \frac{2 sin\; \theta}{sin\; \theta + cos\; \theta} d\theta \\ = \frac{1}{2} \int \frac{(sin\; \theta + cos\; \theta) + (sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ = \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\ = \frac{1}{2} \theta + \frac{1}{2} \int \frac{(sin\; \theta – cos\; \theta)}{(sin\; \theta + cos\; \theta)} d\theta \\$$

$$Suppose,\; (sin\; \theta + cos\; \theta) = z \\ = (cos\; \theta – sin\; \theta) d\theta = dz \\ I = \frac{\theta }{2} + \frac{1}{2} log \left | z \right | + C \\ = \frac{\theta }{2} – \frac{1}{2} log \left | (sin\; \theta + cos\; \theta) \right | + C \\$$

Question 33:

Obtain an integral (or anti – derivative) of the $$\frac{1}{1 – tan \theta}$$

Suppose,

$$\int \frac{1}{1 – tan\; \theta} d\theta \\ = \int \frac{1}{1 – \frac{sin\; \theta}{cos\; \theta}} d\theta \\ = \int \frac{cos\; \theta}{cos\; \theta – sin\; \theta} d\theta \\ = \frac{1}{2} \int \frac{2 cos\; \theta}{cos\; \theta – sin\; \theta} d\theta \\ = \frac{1}{2} \int \frac{(cos\; \theta – sin\; \theta ) + (cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\ = \frac{1}{2} \int 1 d\theta + \frac{1}{2} \int \frac{(cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\ = \frac{1}{2} \theta + \frac{1}{2} \int \frac{(cos\; \theta + sin\; \theta)}{(cos\; \theta – sin\; \theta)} d\theta \\$$ $$Suppose,\; (cos\; \theta – sin\; \theta) = z \\ = (- sin\; \theta – cos\; \theta ) d\theta = dz \\ I = \frac{\theta }{2} – \frac{1}{2} log \left | z \right | + C \\ = \frac{\theta }{2} – \frac{1}{2} log \left | (cos\; \theta – sin\; \theta) \right | + C \\$$

Question 34:

Obtain an integral (or anti – derivative) of the $$\frac{\sqrt{tan \theta}}{sin\; \theta\; cos\; \theta}$$

Answer: Based on formulae given in Integrals

Suppose, $$Suppose, I = \frac{\sqrt{tan \theta}}{sin\; \theta\; cos\; \theta} d\theta \\ = \frac{\sqrt{tan \theta} \times cos\; \theta}{sin\; \theta\; cos\; \theta \times cos\; \theta} d\theta \\ = \int \frac{\sqrt{tan \theta}}{tan\; \theta\; cos\; ^{2} \theta} d\theta \\ = \int \frac{sec\; ^{2} \theta}{\sqrt{tan\; \theta}} d\theta \\ Suppose, tan\; \theta = z \\ sec\; ^{2} \theta d\theta = dz \\ I = \int \frac{dz}{\sqrt{z}} \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{tan\; \theta} + C$$

Question 35:

Obtain an integral (or anti – derivative) of the $$\frac{(1 + log u) ^{2}}{u}$$

Suppose, $$Suppose, 1 + log u = z \\ \frac{1}{u} du = dz \\ \int \frac{(1 + log u) ^{2}}{u} du = \int z ^{2}\; dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{(1 + log u) ^{3}}{3} + C$$

Question 36:

Obtain an integral (or anti – derivative) of the $$\frac{(u + 1)(u + log u) ^{2}}{u}$$

$$\frac{(u + 1)(u + log u) ^{2}}{u} = \frac{(u + 1)}{u} (u + log u) ^{2} = (1 + \frac{1}{u}) (u + log u) ^{2} \\ Suppose,\; (u + log u) = z \\ (1 + \frac{1}{u}) du = dz \\ \int (1 + \frac{1}{u}) (u + log\; u) ^{2} du = \int z ^{2} dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{1}{3} (u + log u) ^{3} + C$$

Question 37:

Obtain an integral (or anti – derivative) of the $$\frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}}$$

$$Suppose,\; u ^{4} = z \\ 4 u ^{3} du = dz \\ \int \frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}} du = \frac{1}{4} \int \frac{sin\; (tan ^{- 1} z)}{1 + z ^{2}} \;dz …. (1) \\ Suppose,\; tan ^{- 1} z = s \\ \frac{1}{1 + z ^{2}} dz = ds \\ From\; (1),\; we\; get\; \\ \int \frac{u ^{3}\; sin\; (tan ^{- 1} u ^{4})}{1 + u ^{8}} du = \frac{1}{4} \int sin\; s\; ds \\ = \frac{1}{4} (- cos s) + c \\ = – \frac{1}{4} cos (tan ^{- 1} z) + C \\ = – \frac{1}{4} cos (tan ^{- 1} u ^{4}) + C \\$$

Question 38:

Which of the following below is the answer for $$\int \frac{10 u ^{9} + 10 ^{u} log_{e} 10}{u ^{10} + 10 ^{u}} du$$ :

$$(a) 10 ^{u} – u ^{10} + C \\ (b) 10 ^{u} + u ^{10} + C \\ (c) (10 ^{u} – u ^{10}) ^{- 1} + C \\ (d) log (10 ^{u} + u ^{10}) + C \\$$

$$u ^{10} + 10 ^{u} = z \\ (10 u ^{9} + 10 ^{u} log_{e} 10) du = dz \\ \int \frac{10 u ^{9} + 10 ^{u} log_{e} 10}{u ^{10} + 10 ^{u}} du = \int \frac{dz}{z} \\ = log z + C \\ = log (u ^{10} + 10 ^{u}) + C \\ Therefore,\; D\; is\; the\; correct\; answer$$

Question 39:

Which of the following below is the answer for $$\int \frac{du}{sin ^{2} \;u\; cos ^{2} \;u}$$

$$(a) tan\; u + cot\; u + C \\ (b) tan\; u – cot\; u + C \\ (c) tan\; u \; cot\; u + C \\ (d) tan\; u – cot\; 2u + C \\$$

$$I = \int \frac{du}{sin ^{2} \;u\; cos ^{2} \;u} \\ = \int \frac{1}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int \frac{sin ^{2} \;u\; + cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int \frac{sin ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du + \int \frac{cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} du \\ = \int sec ^{2} \;u du + \int cosec ^{2} \;u du \\ = tan\; u – cot\; u + C \\$$

Therefore, B is the correct answer

Exercise 7.3

Question 1:

Obtain an integral (or anti – derivative) of the sin 2 (2u + 5)

$$sin ^{2} (2u + 5) = \frac{1 – cos\; 2 (2u + 5)}{2} = \frac{1 – cos\; (4u + 10)}{2} \\ \int sin ^{2} (2u + 5) du = \int \frac{1 – cos\; (4u + 10)}{2} du \\ = \frac{1}{2} \int 1 du – \frac{1}{2} \int cos\; (4u + 10) du \\ = \frac{1}{2} u – \frac{1}{2} \frac{sin (4u + 10)}{4} + C \\ = \frac{1}{2} u – \frac{1}{8} [sin (4u + 10)] + C$$

Question 2:

Obtain an integral (or anti – derivative) of the sin 3u. cos 4u

As we know, $$sin\; C cos\; D = \frac{1}{2} [sin\; (C + D) + sin\; (C – D)] \\ \int sin\; 3u. cos\; 4u\; du = \int \frac{1}{2} [sin\; (3u + 4u) + sin\; (3u – 4u)] \\ = \int \frac{1}{2} [sin\; (7u) + sin\; (- u)] du \\ = \int \frac{1}{2} [sin\; (7u) – sin\; (u)] du \\ = \frac{1}{2} \int sin\; (7u)\; du – \frac{1}{2} \int sin\; (u)\; du \\ = \frac{1}{2} (\frac{- cos\; 7u}{7}) – \frac{1}{2} (- cos\; u) + C \\ = \frac{- cos\; 7u}{14} + \frac{cos\; u}{2} + C$$

Question 3:

Obtain an integral (or anti – derivative) of the cos 2u cos 4u cos 6u

Answer 3: Based on formulae given in Integrals

As we know, $$cos\; C\; cos\; D = \frac{1}{2} [cos\; (C + D) + cos\; (C – D)] \\ \int cos\; 2u\; (cos\; 4u\; cos\; 6u) du = \int cos\; 2u\; [\frac{1}{2} (cos\; (4u + 6u) + cos\; (4u – 6u))] du \\ \frac{1}{2} \int [cos\; 2u\; (cos\; (10 u) + cos\; (- 2u))] du \\ \frac{1}{2} \int [cos\; 2u\; cos\; 10 u + cos\; 2u\; cos\; (- 2u)] du \\ \frac{1}{2} \int [cos\; 2u\; cos\; 10 u + cos ^{2} \;2u]\; du \\ \frac{1}{2} \int [\frac{1}{2}\; (cos\; (2u + 10u) + cos\; (2u – 10u)) + (\frac{1 + cos\; 4u}{2})]\; du \\ \frac{1}{4} \int (cos\; 12u + cos\; 8u + 1 + cos\; 4u)\; du \\ \frac{1}{4} [\frac{sin 12u}{12} + \frac{sin 8u}{8} + u + \frac{sin 4u}{4}] + C \\$$

Question 4:

Obtain an integral (or anti – derivative) of the $$sin ^{3} (2u + 1)$$

$$I = \int sin ^{3} (2u + 1) du \\ \int sin ^{3} (2u + 1) du = \int sin ^{2} \;(2u + 1). sin \;(2u + 1)\; du \\ Supose,\; cos\; (2u + 1) = z \\ – 2 sin (2u + 1)\; du = dz \\ sin (2u + 1)\; du = \frac{- dz}{2} \\ I = \frac{- 1}{2} \int (1 – z^{2}) dz \\ = \frac{- 1}{2} \left \{ z – \frac{z ^{3}}{3} \right \} + C \\ = \frac{- 1}{2} \left \{cos\; (2u + 1) – \frac{cos ^{3} \;(2u + 1)}{3} \right \} + C \\ = \frac{- cos\; (2u + 1)}{2} + \frac{cos ^{3} \;(2u + 1)}{6} + C \\$$

Question 5:

Obtain an integral (or anti – derivative) of the $$sin^{3} \;u\; cos^{3} \;u$$

$$I = \int sin^{3} \;u\; cos^{3} \;u\; du \\ = \int cos^{3} \;u\;. sin^{2} \;u\; sin\; u\; du \\ = \int cos^{3} \;u\; (1 – cos^{2} \;u). sin\; u\; du \\ Suppose,\; cos\; u = z \\ – sin\; u\; du = dz \\ I = – \int z^{3} (1 – z ^{2}) dz \\ = – \int (z^{3} – z ^{5}) dz \\ = – \left \{ \frac{z ^{4}}{4} – \frac{z ^{6}}{6} \right\} + C \\ = – \left \{ \frac{cos ^{4}}{4} – \frac{cos ^{6}}{6} \right\} + C \\ = \frac{cos ^{6}}{6} – \frac{cos ^{4}}{4} + C$$

Question 6:

Obtain an integral (or anti – derivative) of the sin u sin 2u sin 3u

Answer 6: Based on formulae given in Integrals

$$sin\; C\; sin\; D = \frac{1}{2} [cos\; (C – D) – cos\; (C + D)] \\ \int sin\; u\; sin\; 2u\; sin\; 3u\; du = \int sin\; u\;. \frac{1}{2} \left \{ cos\; (2u – 3u) – cos\; (2u + 3u)\right \}\; du \\ = \frac{1}{2} \int (sin\; u\; cos\; (- u) – sin\; u\; cos\; 5u) \; du \\ = \frac{1}{2} \int (sin\; u\; cos\; u – sin\; u\; cos\; 5u) \; du \\ = \frac{1}{2} \int \frac{(2 sin\; u\; cos\; u)}{2} \;du – \frac{1}{2} \int sin\; u\; cos\; 5u \;du \\ = \frac{1}{2} \int \frac{(sin\; 2u)}{2} \;du – \frac{1}{2} \int sin\; u\; cos\; 5u \;du \\$$ $$= \frac{1}{4} \left [ \frac{- cos\; 2u}{2} \right ] – \frac{1}{2} \int \left \{ \frac{1}{2} (sin\; (u + 5u) + sin\; (u – 5u)) \right \} du \\ = \frac{- cos\; 2u}{8} – \frac{1}{4} \int \left \{(sin\; (6u) + sin\; (- 4u)) \right \} du \\ = \frac{- cos\; 2u}{8} – \frac{1}{4} \left [ \frac{- cos\; 6u}{6} + \frac{cos\; 4u}{4} \right ] + C \\ = \frac{- cos\; 2u}{8} – \frac{1}{8} \left [ \frac{- cos\; 6u}{3} + \frac{cos\; 4u}{2} \right ] + C \\ = \frac{1}{8} \left [ \frac{- cos\; 6u}{3} – \frac{cos\; 4u}{2} – cos\; 2u \right ] + C$$

Question 7:

Obtain an integral (or anti – derivative) of the sin 4u sin 8u

As we know, $$sin\; C\; sin\; D = \frac{1}{2} [cos\; (C – D) – cos\; (C + D)] \\ \int sin\; 4u\; sin\; 8u\; du = \int \frac{1}{2} [cos\; (4u – 8u) – cos\; (4u + 8u)]\; du \\ = \frac{1}{2} \int (cos\; (- 4u) – cos\; 12u)\; du \\ = \frac{1}{2} \int (cos\; 4u – cos\; 12u)\; du \\ = \frac{1}{2} \left [ \frac{sin\; 4u}{4} – \frac{sin\; 12u}{12} \right ]$$

Question 8:

Obtain an integral (or anti – derivative) of the $$\frac{1 – cos\; u}{1 + cos\; u}$$

$$\frac{1 – cos\; u}{1 + cos\; u} = \frac{2 sin^{2}\; \frac{u}{2}}{2 cos^{2}\; \frac{u}{2}} \\ = tan^{2}\; \frac{u}{2} \\ = (sec^{2}\; \frac{u}{2} – 1) \\ \int \frac{1 – cos\; u}{1 + cos\; u} du = \int (sec^{2}\; \frac{u}{2} – 1) \;du \\ = \left [ \frac{tan\; \frac{u}{2}}{\frac{1}{2}} – u \right ] + C \\ = 2 tan\; \frac{u}{2} – u + C$$

Question 10:

Obtain an integral (or anti – derivative) of the sin 4 u.

Answer 10: Based on formulae given in Integrals

$$sin ^{4} \;u = sin ^{2} \;u \times sin ^{2} \;u \\ = (\frac{1 – cos \;2u}{2}) (\frac{1 – cos \;2u}{2}) \\ = \frac{1}{4} (1 – cos \;2u)^{2} \\ = \frac{1}{4} (1 + cos^{2} \;2u – 2\; cos \;2u) \\ = \frac{1}{4} \left [ 1 + (\frac{1 + cos\; 4u}{2}) – 2 cos\; 2u \right ] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u] \\ = \frac{1}{4} \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u\right ] \\$$ $$\int sin ^{4} \;u\; du = \frac{1}{4}\; \int \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 4u – 2 cos\; 2u\right ] du \\ = \frac{1}{4}\; \left [ \frac{3}{2} u + \frac{1}{2}\; (\frac{sin\; 4u}{4}) – sin\; 2u \right ] + C \\ = \frac{3}{8} u + (\frac{sin\; 4u}{32}) – \frac{sin\; 2u}{4} + C$$

Question 11:

Obtain an integral (or anti – derivative) of the cos 4 2u

$$cos ^{4} \;2u = (sin ^{2} \;2u) ^{2} \\ = (\frac{1 + cos \;4u}{2}) ^{2} \\ = \frac{1}{4} (1 + cos^{2} \;4u + 2\; cos \;4u) \\ = \frac{1}{4} \left [ 1 + (\frac{1 + cos\; 8u}{2}) + 2 cos\; 4u \right ] \\ = \frac{1}{4} [1 + \frac{1}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u] \\ = \frac{1}{4} \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u\right ] \\$$ $$\int cos ^{4} \;u\; du = \frac{1}{4}\; \int \left [ \frac{3}{2} + \frac{1}{2}\; cos\; 8u + 2 cos\; 4u\right ] du \\ = \frac{1}{4}\; \left [ \frac{3}{2} u + \frac{1}{2}\; (\frac{sin\; 8u}{4}) + sin\; 4u \right ] + C \\ = \frac{3}{8} u + (\frac{sin\; 8u}{64}) + \frac{sin\; 4u}{8} + C$$

Question 12:

Obtain an integral (or anti – derivative) of the $$\frac{sin ^{2} \;u}{1 + cos\; u}$$

$$\frac{sin ^{2} \;u}{1 + cos\; u} = \frac{(2 sin\; \frac{u}{2}\; cos\; \frac{u}{2}) ^{2}}{2\; cos^{2} \frac{u}{2}} \;\;\;[Since\; sin\; u = 2\; sin\; \frac{u}{2}\; cos\; \frac{u}{2}; \; cos\; u = 2\; cos^{2} \frac{u}{2} – 1] \\ = \frac{4 sin ^{2} \;\frac{u}{2}\; cos ^{2} \;\frac{u}{2}}{2\; cos^{2} \frac{u}{2}} \\ = 2 sin ^{2} \;\frac{u}{2} \\ = 1 – cos\; u \\ \int \frac{sin ^{2} \;u}{1 + cos\; u} du = \int (1 – cos\; u) \; du \\ = u – sin\; u + C$$

Question 13:

Obtain an integral (or anti – derivative) of the $$\frac{cos\; 2u – cos\; 2a}{cos\; u – cos\; a}$$

Answer 13: Based on formulae given in Integrals

$$\frac{cos\; 2u – cos\; 2a}{cos\; u – cos\; a} = \frac{- 2\; sin \;\frac{2u + 2a}{2}\; sin \;\frac{2u – 2a}{2}}{- 2\; sin \;\frac{u + a}{2}\; sin \;\frac{u – a}{2}} \; \; \left [Since,\; cos\; A – cos\; B = – 2\; sin\; \frac{A + B}{2}\; sin\; \frac{A – B}{2} \right ] \\ = \frac{sin\; (u + a)\; sin\; (u – a)}{ sin \;\frac{u + a}{2}\; sin \;\frac{u – a}{2}} \\ = \frac{\left [ 2 sin\; \frac{u + a}{2}\; cos\; \frac{u + a}{2} \right ] \left [ 2 sin\; \frac{u – a}{2}\; cos\; \frac{u – a}{2} \right ]}{sin\; \frac{u + a}{2}\; sin\; \frac{u – a}{2}} \\$$ $$= 4 cos\; \frac{u + a}{2}\; cos\; \frac{u – a}{2} \\ = 2 \left [ cos\; \left ( \frac{u + a}{2} + \frac{u – a}{2} \right ) + cos\; \left ( \frac{u + a}{2} – \frac{u – a}{2} \right ) \right ] \\ = 2 [cos\; (u) + cos\; a] \\ = 2 cos\; u + 2 cos\; a$$

Question 14:

Obtain an integral (or anti – derivative) of the $$\frac{cos\; u – sin\; u}{1 + sin\; 2u}$$

$$\frac{cos\; u – sin\; u}{1 + sin\; 2u} = \frac{cos\; u – sin\; u}{(sin ^{2} \;u + cos ^{2} \;u) + 2 sin\; u\; cos\; u} \\ Since,\; sin ^{2} \;u + cos ^{2} \;u = 1; sin\; 2u = 2 sin\; u\; cos\; u \\ = \frac{cos\; u – sin\; u}{(sin\; u + cos\; u) ^{2}} \\ Suppose,\; sin\; u + cos\; u = z \\ (cos\; u – sin\; u)\; du = dz \\ \int \frac{cos\; u – sin\; u}{1 + sin\; 2u}\; du = \int \frac{cos\; u – sin\; u}{(sin\; u + cos\; u) ^{2}} du \\$$ $$= \frac{dz}{z ^{2}} \\ = \int z ^{- 2} dz \\ = – z ^{- 1} + C \\ = – \frac{1}{z} + C \\ = \frac{- 1}{sin\; u + cos\; u} + C$$

Question 15:

Obtain an integral (or anti – derivative) of the $$tan ^{3}\; 2u\; sec\; 2u$$

$$tan ^{3}\; 2u\; sec\; 2u = tan ^{2}\; 2u\; tan\; 2u\; sec\; 2u \\ = (sec ^{2} \;2u – 1) tan\; 2u\; sec\; 2u \\ = sec ^{2} \;2u\; tan\; 2u\; sec\; 2u – tan\; 2u\; sec\; 2u \\ = \int tan ^{3}\; 2u\; sec\; 2u du = \int sec ^{2} \;2u\; tan\; 2u\; sec\; 2u\; du – \int tan\; 2u\; sec\; 2u \;du \\ = \int sec ^{2} \;2u\; tan\; 2u\; sec\; 2u\; du – \frac{sec\; 2u}{2} + C \\$$ $$Suppose,\; sec\; 2u = z \\ 2 sec\; 2u\; tan\; 2u du = dz \\ \int tan ^{3}\; 2u\; sec\; 2u = \frac{1}{2} z ^{2} dz – \frac{sec\; 2u}{2} + C \\ = \frac{z ^{3}}{6} – \frac{sec\; 2u}{2} + C \\ = \frac{(sec\; 2u) ^{3}}{6} – \frac{sec\; 2u}{2} + C$$

Question 16:

Obtain an integral (or anti – derivative) of the $$tan ^{4} \;u$$

Answer 16: Based on formulae given in Integrals

$$tan ^{4} \;u = tan ^{2} \;u\; \times tan ^{2} \;u \\ = (sec ^{2} \;u\; – 1) tan ^{2} \;u \\ = sec ^{2} \;u\; tan ^{2} \;u – tan ^{2} \;u \\ = sec ^{2} \;u\; tan ^{2} \;u – (sec ^{2} \;u\; – 1) \\ = sec ^{2} \;u\; tan ^{2} \;u – sec ^{2} \;u\; + 1 \\$$ $$\int tan ^{4} \;u\; du = \int sec ^{2} \;u\; tan ^{2} \;u\; du – \int sec ^{2} \;u\; du + \int 1\; du \\ = \int sec ^{2} \;u\; tan ^{2} \;u\; du – tan\; u + u + C \;\; …… (1) \\ Now,\; \int sec ^{2} \;u\; tan ^{2} \;u\; du \\ Suppose,\; tan\; u = z \\ sec ^{2} \;u\; du = dz \\ \int sec ^{2} \;u\; tan ^{2} \;u\; du = \int z ^{2}\; dz \\ = \frac{z ^{3}}{3} + C \\ = \frac{tan ^{3} \;u}{3} \\ Therefore, from equation (1) is\\ \int tan ^{4} \;u\; du = \frac{tan ^{3} \;u}{3} – tan\; u + u + C$$

Question 17:

Obtain an integral (or anti – derivative) of the $$\frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u}$$

$$\frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} = \frac{sin ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} + \frac{cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u} = \frac{sin \;u}{cos ^{2} \;u} + \frac{cos \;u}{sin ^{2} \;u} \\$$ $$= tan\; u\; sec\; u + cot\; u\; cosec\; u \\ Therefore,\; \int \frac{sin ^{3} \;u + cos ^{3} \;u}{sin ^{2} \;u\; cos ^{2} \;u}\; du = \int (tan\; u\; sec\; u + cot\; u\; cosec\; u)\; du \\ = sec\; u – cosec\; u + C \\$$

Question 18:

Obtain an integral (or anti – derivative) of the $$\frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u} \\$$

$$\frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u} \\ \frac{cos\; 2u + (1 – cos\; 2u)}{cos ^{2} \;u}\; \; [Since,\; cos\; 2u = 1 – 2 sin ^{2} \;u] \\ = \frac{1}{cos ^{2} \;u} \\ = sec ^{2} \;u \\ \int \frac{cos\; 2u + 2 sin ^{2} \;u}{cos ^{2} \;u}\; du = \int sec ^{2} \;u\; du = tan\; u + C$$

Question 19:

Obtain an integral (or anti – derivative) of the $$\frac{1}{sin\; u\; cos^{2} \;u}$$

Answer 19: Based on formulae given in Integrals

$$\frac{1}{sin\; u\; cos^{2} \;u} = \frac{sin^{2} \;u + cos^{2} \;u}{sin\; u\; cos^{2} \;u} \\ = \frac{sin\; u\;}{cos^{2} \;u} + \frac{1}{sin\; u\; cos\; u} \\ = tan\; u\; sec^{2} \;u + \frac{1}{sin\; u\; cos\; u} \times \frac{cos ^{2} \;u}{cos ^{2} \;u} \\ = tan\; u\; sec^{2} \;u + \frac{sec ^{2} \;u}{tan\; u}$$ $$\int \frac{1}{sin\; u\; cos^{2} \;u} du = \int tan\; u\; sec^{2} \;u\; du + \int \frac{sec ^{2} \;u}{tan\; u}\; du \\$$ $$Suppose,\; tan\; u = z \\ sec ^{2} \;u\; du = dz \\ \int \frac{1}{sin\; u\; cos^{2} \;u}\; du = \int z\; dz + \int \frac{1}{z}\; dz\; \\ = \frac{z ^{2}}{2} + log \left | z \right | + C \\ = \frac{1}{2} tan^{2} \;u + log \left | tan\; u \right | + C$$

Question 20:

Obtain an integral (or anti – derivative) of the $$\frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}$$

$$\frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}} = \frac{cos\; 2u}{cos ^{2} \;u + sin^{2} \;u + 2 sin\; u\; cos\; u} = \frac{cos\; 2u}{1 + sin\; 2u} \\ \int \frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}\; du = \int \frac{cos\; 2u}{(1 + sin\; 2u)}\; du \\ Suppose,\; 1 + sin\; 2u = Z \\ 2 cos\; 2u\; du = dz \\ \int \frac{cos\; 2u}{(cos\; u + sin\; u) ^{2}}\; du = \frac{1}{2} \int \frac{1}{z}\; dz \\ = \frac{1}{2} log \left | z \right | + C \\ = \frac{1}{2} log \left | 1 + sin\; 2u \right | + C \\ = \frac{1}{2} log \left | (cos\; u + sin\; u) ^{2} \right | + C \\ = \frac{1}{2} log \left | cos\; u + sin\; u \right | + C$$

Question 21:

Obtain an integral (or anti – derivative) of the $$sin ^{- 1} (cos\; u)$$

Answer 21: Based on formulae given in Integrals

$$sin ^{- 1} (cos\; u) \\ Suppose,\; cos\; x = z \\ Then, sin\; u = \sqrt{1 – u ^{2}} \\ (- sin\; u)\; du = dz \\ du = \frac{- dz}{\sqrt{1 – z ^{2}}} \\ Therefore,\; \int sin ^{- 1} (cos\; u)\; du = \int sin ^{- 1}\; z (- \frac{dz}{\sqrt{1 – z ^{2}}})\; dz \\ Suppose,\; sin ^{- 1} \;z = p \\ \frac{1}{\sqrt{1 – z ^{2}}} \;dz = dp \\$$ $$Therefore,\; \int sin ^{- 1} (cos\; u)\; du = – \int p\; dp \\ = – \frac{p ^{2}}{2} + C \\ = – \frac{(sin ^{- 1}\; z) ^{2}}{2} + C \\ = – \frac{(sin ^{- 1}\; (cos\; u)) ^{2}}{2} + C ….. (1) \\ As\; we\; know, \\ sin ^{- 1} \;u + cos ^{- 1} \;u\; = \frac{\pi }{2} \\ Therefore,\; sin ^{- 1} (cos\; u) = \frac{\pi }{2} – cos ^{- 1} (cos\; u) = (\frac{\pi }{2} – u) \\ On\; substituting\; in\; equation\; (1), we\; get,\;$$ $$\int sin ^{- 1} (cos\; u)\; du = \frac{- \left [ \frac{\pi}{2} – u \right ] ^{2}}{2} + C \\ = – \frac{1}{2}\; \left ( (\frac{\pi }{2}) ^{2} + u^{2} – \pi u \right ) + C \\ = – \frac{(\pi )^{2}}{8} – \frac{u ^{2}}{2} + \frac{1}{2} \pi u + C \\ = \frac{\pi u}{2} – \frac{u ^{2}}{2} + \left ( c – \frac{(\pi )^{2}}{8} \right ) = \frac{\pi u}{2} – \frac{u ^{2}}{2} + C_{1}$$

Question 22:

Obtain an integral (or anti – derivative) of the $$\frac{1}{cos\; (u – m)\; cos\; (u – n)}$$

$$\frac{1}{cos\; (u – m)\; cos\; (u – n)} = \frac{1}{sin\; (m – n)} \left [ \frac{sin\; (m – n)}{cos\; (u – m)\; cos\; (u – n)} \right ] \\ = \frac{1}{sin\; (m – n)} \left [ \frac{sin\; [(u – n) – (u – m)]}{cos\; (u – m)\; cos\; (u – n)} \right ] \\ = \frac{1}{sin\; (m – n)} \frac{sin\; (u – n)\; cos\; (u – m) – cos\; (u – n)\; sin\; (u – m)}{cos\; (u – m)\; cos\; (u – n)} \\ = \frac{1}{sin\; (m – n)} [tan\; (u – n) – tan\; (u – m)]$$ $$\int \frac{1}{cos\; (u – m)\; cos\; (u – n)} du = \frac{1}{sin\; (m – n)} \int [tan\; (u – n) – tan\; (u – m)] du \\ = \frac{1}{sin\; (m – n)} [- log \left | cos\; (u – n) \right | + log \left | cos\; (u – m) \right |] \\ = \frac{1}{sin\; (m – n)} log\; \left [ \frac{cos\; (u – m)}{cos\; (u – n)} \right ] + C \\$$

Question 23:

Which of the following below is the answer for $$\frac{sin ^{2} \;u – cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u}$$

(a) tan u + cot u + C

(b) tan u + cosec u + C

(c) – tan u + cot u + C

(d) tan u + sec u + C

Answer 23: Based on formulae given in Integrals

$$\int \frac{sin ^{2} \;u – cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u}\; du \\ = \int \left ( \frac{sin ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} – \frac{cos ^{2} \;u}{sin ^{2} \;u\; cos ^{2} \;u} \right )\; du \\ = \int (sec ^{2}\; u\; – cosec ^{2}\; u)\; du \\ = tan\; u + cot\; u + C$$

Thus, (a) is the correct answer.

Question 24:

Which of the following below is the answer for $$\int \frac{e ^{u} (1 + u)}{cos ^{2}\; (e ^{u} u)}\; du$$

(a) – cot (e uu) + C

(b) tan (e uu) + C

(c) tan (eu) + C

(d) cot (eu) + C

$$\int \frac{e ^{u} (1 + u)}{cos ^{2}\; (e ^{u} u)}\; du \\ Suppose,\; e ^{u} u = z \\ (e ^{u}. u + e ^{u}. 1) du = \int \frac{dz}{cos ^{2}\; z} \\ = \int sec ^{2} \;z dz \\ = tan\; z + C \\ = tan (e ^{u}. u) + C$$

Thus, (b) is the correct answer.

Exercise 7.4

Question 1:

Obtain an integral (or anti – derivative) of the $$\frac{3 u^{2}}{u^{6} + 1}$$

$$Suppose,\; u ^{3} = z \\ 3 u ^{2} du = dz \\ \int \frac{3 u^{2}}{u^{6} + 1} du = \int \frac{dz}{z ^{2} + 1} \\ = tan ^{- 1}\; z + C \\ = tan ^{- 1}\; u ^{3} + C$$

Question 2:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{1 + 4 u ^{2}}}$$

$$Suppose,\; 2 u = z \\ 2\; du = dz \\ \int \frac{1}{\sqrt{1 + 4 u ^{2}}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{1 + z ^{2}}} \\ = \frac{1}{2} [log \left | z + \sqrt{1 + z ^{2}} \right |] + C \\ = \frac{1}{2} [log \left | 2 u + \sqrt{1 + 4 u^{2}} \right |] + C$$

Question 3:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{(2 – u)^{2} + 1}}$$

Answer 3: Based on formulae given in Integrals

$$Suppose,\; 2 – u = z \\ – du = dz \\ \int \frac{1}{\sqrt{(2 – u)^{2} + 1}}\; du = – \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = – [log \left | z + \sqrt{z ^{2} + 1} \right |] + C \\ = – [log \left | 2 – u + \sqrt{(2 – u) ^{2} + 1} \right |] + C \\ = log \left | \frac{1}{(2 – u) + \sqrt{u ^{2} – 4u + 5}} \right | + C$$

Question 4:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{9 – 25 u^{2}}}$$

Suppose, 5u = z

5 du = dz

$$\int \frac{1}{\sqrt{9 – 25 u^{2}}}\; du = \frac{1}{5} \int \frac{1}{9 – z ^{2}}\; dz \\ = \frac{1}{5} \int \frac{1}{\sqrt{3 ^{2} – z ^{2}}}\; dz \\ = \frac{1}{5} sin ^{- 1} \;\frac{z}{3} + C \\ = \frac{1}{5} sin ^{- 1} \;\frac{5 u}{3} + C \\$$

Question 5:

Obtain an integral (or anti – derivative) of the $$\frac{3 u}{1 + 2 u ^{4}}$$

Answer 5: Based on formulae given in Integrals

$$Suppose,\; \sqrt{2} u^{2} = z \\ 2 \sqrt{2}\; u\; du = dz \\ \int \frac{3 u}{1 + 2 u ^{4}}\; du = \frac{3}{2 \sqrt{2}} \int \frac{dz}{1 + z ^{2}} dz \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} z] + C \\ = \frac{3}{2 \sqrt{2}} [tan ^{- 1} (\sqrt{2} u^{2})] + C$$

Question 6:

Obtain an integral (or anti – derivative) of the $$\frac{u ^{2}}{1 – u ^{6}}$$

Suppose, u3 = z

3 u2 du = dz

$$\int \frac{u ^{2}}{1 – u ^{6}} du = \frac{1}{3} \int \frac{dz}{1 – z^{2}} \\ = \frac{1}{3} \left [ \frac{1}{2} log \left |\frac{1 + z}{1 – z} \right |\right ] + C \\ = \frac{1}{6} log \left | \frac{1 + u^{3}}{1 – u^{3}} \right | + C$$

Question 7:

Obtain an integral (or anti – derivative) of the $$\frac{u – 1}{u ^{2} – 1}$$

Answer 7: Based on formulae given in Integrals

$$\int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ For, \int \frac{u}{\sqrt{u ^{2} – 1}}\; du \\$$ $$Suppose\; u ^{2} – 1 = z \\ 2u\; du = dz \\ Therefore,\; \int \frac{u}{\sqrt{u ^{2} – 1}}\; du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} \int z ^{- \frac{1}{2}} dz \\ = \frac{1}{2} [2 z^{\frac{1}{2}}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} + C$$ $$From\; the\; above\; equation\; we\; get \\ \int \frac{u – 1}{\sqrt{u ^{2} – 1}} du = \int \frac{u}{\sqrt{u ^{2} – 1}}\; du – \frac{1}{\sqrt{u ^{2} – 1}}\; du \\ = \sqrt{u ^{2} – 1} + C – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{1} \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + (C + C_{1}) \\ = \sqrt{u ^{2} – 1} – log \left | u + \sqrt{u ^{2} – 1} \right | + C_{2}$$

Question 8:

Obtain an integral (or anti – derivative) of the $$\frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}$$

Suppose, u 3 = z

3 u2 du = dz

$$\int \frac{u ^{2}}{\sqrt{u ^{6} + m ^{6}}}\; du = \frac{1}{3} \int \frac{dz}{\sqrt{z ^{2} + (m^{3}) ^{2}}} \\ = \frac{1}{3} log \left | z + \sqrt{z ^{2} + m^{6}} \right | + C \\ = \frac{1}{3} log \left | u ^{3} + \sqrt{u ^{6} + m^{6}} \right | + C$$

Question 9:

Obtain an integral (or anti – derivative) of the $$\frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}$$

Answer 9: Based on formulae given in Integrals

Suppose, tan u = z

sec 2 u du = dz

$$\int \frac{sec ^{2}\; u}{\sqrt{tan ^{2} \;u + 4}}\; du = \int \frac{dz}{\sqrt{z ^{2} + 2 ^{2}}} \\ = log \left | z + \sqrt{z ^{2} + 4} \right | + C \\ = log \left | tan\; u + \sqrt{tan ^{2} \;u + 4} \right | + C$$

Question 10:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{u ^{2} + 2u + 2}}$$

$$\int \frac{1}{\sqrt{u ^{2} + u + 2}} du = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \\ Suppose,\; u + 1 = z \\ du = dz => \int \frac{1}{\sqrt{u ^{2} + 2 u + 2}} du = \int \frac{1}{\sqrt{z ^{2} + 1}} \;dz \\ = log \left | z + \sqrt{z ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{(u + 1) ^{2} + 1} \right | + C \\ = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 1} \right | + C$$

Question 11:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{9 u ^{2} + 6 u + 2}}$$

$$\int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \int \frac{1}{(3 u + 1) ^{2} + (2) ^{2}} \\ Suppose,\; 3 u + 1 = z \\ 3\; du = dz => \int \frac{1}{\sqrt{9 u ^{2} + 6 u + 2}} du = \frac{1}{3} \int \frac{1}{\sqrt{t ^{2} + 2 ^{2}}} \;dz \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{z}{2}) \right ] + C \\ = \frac{1}{3} \left [ \frac{1}{2} tan ^{- 1} (\frac{3 u + 1}{2}) \right ] + C$$

Question 12:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{7 – 6 u – u ^{2}}}$$

Answer 12: Based on formulae given in Integrals

7 – 6 u – u2 = can also be written as 7 – (u2 + 6 u + 9 – 9)

Therefore,

7 – (u2 + 6 u + 9 – 9)

= 16 – (u2 + 6 u + 9)

= 16 – (u + 3) 2

= 4 2 – (u + 3) 2

$$\int \frac{1}{\sqrt{7 – 6 u – u ^{2}}} du = \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du \\ Suppose,\; u + 3 = z \\ du = dz \\ \int \frac{1}{4 ^{2} – (u + 3) ^{2}} du = \int \frac{1}{4 ^{2} – (z) ^{2}} dz \\ = sin ^{- 1} (\frac{z}{4}) + C \\ = sin ^{- 1} (\frac{u + 3}{4}) + C$$

Question 13:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{(u – 1) (u – 2)}}$$

Answer 13: Based on formulae given in Integrals

(u – 1) (u – 2) can be written as u 2 – 3 u + 2

Therefore,

u 2 – 3 u + 2

$$= u ^{2} – 3 u + \frac{9}{4} – \frac{9}{4} + 2 \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2} \\$$ $$\int \frac{1}{\sqrt{(u – 1) (u – 2)}}\; du = \int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du \\ Suppose,\; u – \frac{3}{2} = z \\ du = dz \\$$ $$\int \frac{1}{\sqrt{\left ( u – \frac{3}{2} \right ) ^{2} – \left ( \frac{1}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – (\frac{1}{2}) ^{2}}} dz \\ = log \left | z + \sqrt{z ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{(u – \frac{3}{2}) ^{2} – (\frac{1}{2}) ^{2}} \right | + C \\ = log \left | (u – \frac{3}{2}) + \sqrt{u^{2} – 3 u + 2} \right | + C$$

Question 14:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{8 + 3 u – u^{2}}}$$

Answer 14: Based on formulae given in Integrals

$$\frac{1}{\sqrt{8 + 3 u – u^{2}}} can\; also\; be\; written\; as\; 8 – \left ( u^{2} – 3 u + \frac{9}{4} – \frac{9}{4} \right ) \\ Therefore,\; \frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2} \\ \int \frac{1}{\sqrt{8 + 3 u – u^{2}}}\; du = \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du \\$$ $$Suppose\; u – \frac{3}{2} = z \\ du = dz \\ \int \frac{1}{\sqrt{\frac{41}{4} – \left ( u – \frac{3}{2} \right ) ^{2}}}\; du = \frac{1}{\sqrt{\left ( \frac{\sqrt{41}}{2} \right ) ^{2} – z ^{2} }} \; dz \\ = sin ^{- 1} \left ( \frac{z}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{u – \frac{3}{2}}{\frac{\sqrt{41}}{2}} \right ) + C \\ = sin ^{- 1} \left ( \frac{2 u – 3}{\sqrt{41}} \right ) + C \\$$

Question 15:

Obtain an integral (or anti – derivative) of the $$\frac{1}{\sqrt{(u – m) (u – n)}}$$

Answer 15: Based on formulae given in Integrals

$$(u – m) (u – n)\; can\; also\; be\; written\; as\; u ^{2} – (m + n) u + mn \\ Therefore,\; \\ u ^{2} – (m + n) u + mn \\ = u ^{2} – (m + n) u + \frac{(m + n) ^{2}}{4} – \frac{(m + n) ^{2}}{4} + mn \\ = \left [ u – \frac{(m + n)}{2} \right ] ^{2} – \frac{(m – n) ^{2}}{4} \\ \int \frac{1}{\sqrt{(u – m) (u – n)}}\; du = \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du \\$$ $$Suppose,\; u – \left ( \frac{(m + n)}{2} \right ) = z \\ du = dz \\ \int \frac{1}{\sqrt{\left \{u – \frac{(m + n)}{2} \right \} ^{2} – \left ( \frac{(m – n)}{2} \right ) ^{2}}} \;du = \int \frac{1}{\sqrt{z ^{2} – \left ( \frac{m – n}{2} \right ) ^{2}}} \;dz \\$$ $$log\; \left |z + \sqrt{z ^{2} – (\frac{m – n}{2}) ^{2}} \right | + C \\ log\; \left | \left \{ u – (\frac{m + n}{2}) \right \} + \sqrt{(u – m) (u – n)} \right | + C$$

Question 16:

Obtain an integral (or anti – derivative) of the $$\frac{4 u + 1}{\sqrt{2 u^{2} + u – 3}}$$

Answer 16: Based on formulae given in Integrals

$$Suppose,\; 4u + 1 = A \frac{d}{dx} (2 u^{2} + u – 3) + B …. (1) \\ 4 u + 1 = A (4u + 1) + B \\ 4 u + 1 = 4 Au + A + B$$

Equate the coefficients of u and the constants on both the sides, we get

4 A = 4 => A = 1

A + B = 1 => B = 0

From (1), we get

Suppose, 2 u2 + u – 3 = z

(4 u + 1) du = dz

$$\int \frac{4 u + 1}{\sqrt{2 u^{2} + u – 3}} du = \int \frac{1}{\sqrt{z}} \;dz \\ = 2 \sqrt{z} + C \\ = 2 \sqrt{2 u^{2} + u – 3} + C$$

Question 17:

Obtain an integral (or anti – derivative) of the $$\frac{u + 2}{\sqrt{u ^{2} – 1}}$$

Answer 17: Based on formulae given in Integrals

$$Suppose,\; u + 2 = A \frac{d}{du} (u ^{2} – 1) + B …. (1) \\ u + 2 = A (2 u) + B$$

Equate the coefficients of u and the constants on both the sides, we get

2 A = 1 => A = (1 / 2)

B = 2

From (1), we get

$$From\; (1), we\; get, \\ (u + 2) = \frac{1}{2} (2 u) + 2 \\ Then,\; \int \frac{u + 2}{\sqrt{u ^{2} – 1}}\; du = \int \frac{\frac{1}{2} (2 u) + 2}{\sqrt{u ^{2} – 1}} \\ = \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du + \int \frac{2}{\sqrt{u ^{2} – 1}} \;du ….. (2) \\$$ $$In\; \frac{1}{2} \int \frac{2 u}{\sqrt{u ^{2} – 1}} \;du = \frac{1}{2} \int \frac{dz}{\sqrt{z}} \\ = \frac{1}{2} [2 \sqrt{z}] + C \\ = \sqrt{z} + C \\ = \sqrt{u ^{2} – 1} \\ Then,\; \int \frac{2}{\sqrt{u ^{2} – 1}} \;du = 2 \int \frac{1}{\sqrt{u ^{2} – 1}} \;du \\ In\; equation\; (2), we\; get, \\ \int \frac{u + 2}{\sqrt{u ^{2} – 1}} \;du = \sqrt{u ^{2} – 1} + 2\; log\; \left | u + \sqrt{u ^{2} – 1} \right | + C$$

Question 18:

Obtain an integral (or anti – derivative) of the $$\frac{5u – 2}{1 + 2 u + 3 u^{2}}$$

Answer 18: Based on formulae given in Integrals

Suppose, $$5 u – 2 = A \frac{d}{du} (1 + 2 u + 3 u^{2}) + B \\ 5 u – 2 = A (2 + 6 u) + B \\$$

Equate the coefficients of u and the constants on both the sides, we get

$$5 = 6 A => A = \frac{5}{6} \\ 2 A + B = – 2 => B = – \frac{11}{3} \\ 5 u – 2 = \frac{5}{6} (2 + 6 u) + \frac{- 11}{3} \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \int \frac{\frac{5}{6} (2 + 6 u) – \frac{11}{3}}{1 + 2 u + 3 u^{2}} \;du \\ = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\$$ $$Suppose,\; I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \; and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} I_{1} – \frac{11}{3} I_{2} …. (1) \\ I_{1} = \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du \;and\; I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ Suppose,\; 1 + 2 u + 3 u^{2} = z \\ (2 + 6u)\; du = dz \\ I_{1} = \int \frac{dz}{z} \\ I_{1} = log \left | z \right | + C \\ I_{1} = log \left | 1 + 2 u + 3 u^{2} \right | + C ….. (2) I_{2} = \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\$$ $$1 + 2 u + 3 u^{2} can\; also\; be\; written\; as\; 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ Therefore, \; \\ 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u \right ) \\ = 1 + 3 \left ( u ^{2} + \frac{2}{3}\; u + \frac{1}{9} – \frac{1}{9} \right ) \\ = 1 + 3 \left ( u + \frac{1}{3} \right ) ^{2} – \frac{1}{3} \\ = \frac{2}{3} + 3 \left ( u + \frac{1}{3} \right ) ^{2} \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + \frac{2}{9}] \\ = 3 [\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}] \\$$ $$I_{2} = \frac{1}{3} \int \left [ \frac{1}{[\left ( u + \frac{1}{3} \right ) ^{2} + (\frac{\sqrt{2}}{3}) ^{2}]} \right ] du \\ = \frac{1}{3} \left [ \frac{1}{\frac{\sqrt{2}}{3}}\; tan ^{- 1} (\frac{u + \frac{1}{3}}{\frac{\sqrt{2}}{3}}) \right ] \\ = \frac{1}{3} \left [ \frac{3}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C ….. (3) \\$$

Substituting equations (2) and (3) in equation (1), we get,

$$\int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \int \frac{2 + 6 u}{1 + 2 u + 3 u^{2}} \;du – \frac{11}{3} \int \frac{1}{1 + 2 u + 3 u^{2}} \;du \\ \int \frac{5u – 2}{1 + 2 u + 3 u^{2}} \;du = \frac{5}{6} \left [ log \left | 1 + 2 u + 3 u^{2} \right | \right ] – \frac{11}{3} \left [\frac{1}{\sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) \right ] + C \\ = \frac{5}{6} log \left | 1 + 2 u + 3 u^{2} \right ] – \frac{11}{3 \sqrt{2}} tan ^{- 1} (\frac{3 u + 1}{\sqrt{2}}) + C$$

Question 19:

Obtain an integral (or anti – derivative) of the $$\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}}$$

Answer 19: Based on formulae given in Integrals

Suppose, $$\frac{6 u + 7}{\sqrt{(u – 5) (u – 4)}} = \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \\ Suppose,\; 6 u + 7 = A \frac{d}{du} (u ^{2} – 9 u + 20) + B \\ 6 u + 7 = A (2 u – 9) + B$$

Equate the coefficients of u and the constants on both the sides, we get,

2 A = 6 => A = 3

– 9 + B = 7 => B = 34

6 u + 7 = 3 (2 u – 9) + 34

$$\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = \int \frac{3 (2 u – 9) + 34}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ = 3 \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du + 34 \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du$$ $$Suppose,\; I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \;and \;I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ \int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 I_{1} + 34 I_{2} \\ I_{1} = \int \frac{2 u – 9}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ Suppose,\; u ^{2} – 9 u + 20 = z (2 u – 9) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} \\ I_{1} = 2 \sqrt{z} \\ I_{1} = 2 \sqrt{u ^{2} – 9 u + 20} \\ and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} – 9 u + 20}} \;du \\ u ^{2} – 9 u + 20 \;can\; also \;be\; written\; as\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4}$$ $$Therefore,\; u ^{2} – 9 u + 20 + \frac{81}{4} – \frac{81}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – \frac{1}{4} \\ = \left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{\left ( u – \frac{9}{2} \right ) ^{2} – (\frac{1}{2}) ^{2}}} \;du \\ I_{2} = log \left |(u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right | + C ….. (3)$$

Substituting equations (2) and (3) in equation (1), we get,

$$\int \frac{6 u + 7}{\sqrt{u ^{2} – 9 u + 20}} \;du = 3 [2 \sqrt{u ^{2} – 9 u + 20}] + 34\; log \left [ \left ( (u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C \\ = 6 \sqrt{u ^{2} – 9 u + 20} + 34\; log \left [ \left ((u – \frac{9}{2}) + \sqrt{u ^{2} – 9 u + 20} \right ) \right ] + C$$

Question 20:

Obtain an integral (or anti – derivative) of the $$\frac{u + 2}{\sqrt{4u – u ^{2}}}$$

Answer 20: Based on formulae given in Integrals

Suppose, $$u + 2 = A \frac{d}{du} (4u – u ^{2}) + B \\ (u + 2) = A (4 – 2 u) + B$$

Equate the coefficients of u and the constants on both the sides, we get,

$$– 2 A = 1 => A = – \frac{1}{2} \\ 4 A + B = 2 => B = 4 \\ (u + 2) = – \frac{1}{2} (4 – 2u) + 4 \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = \int \frac{- \frac{1}{2}\; (4 – 2u) + 4}{\sqrt{4 u – u^{2}}} \;du \\ = – \frac{1}{2} \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du + 4 \int \frac{1}{\sqrt{4 u – u^{2}}} \;du$$ $$Suppose,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du\; and\; I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ \int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} I_{1} + 4 I_{2} ….. (1) \\ Then,\; I_{1} = \int \frac{4 – 2 u}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u ^{2} = z \\ (4 – 2 u) \;du = dz \\ I_{1} = \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2\; \sqrt{4u – u ^{2}} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{4 u – u^{2}}} \;du \\ Suppose,\; 4 u – u^{2} = – (- 4 u + u^{2}) \\ (4 – 2 u) \;du = – (- 4 u + u^{2} + 4 – 4) \\ = 4 – (u – 2) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(2) ^{2} – (u – 2) ^{2}}} \;du = sin ^{- 1} (\frac{u – 2}{2}) ….. (3) \\$$

Substituting equations (2) and (3) in equation (1), we get,

$$\int \frac{u + 2}{\sqrt{4u – u ^{2}}} \;du = – \frac{1}{2} (2 \sqrt{4 u – u ^{2}}) + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C \\ = – \sqrt{4u – u ^{2}} + 4 sin^{- 1} \left ( \frac{u – 2}{2} \right ) + C$$

Question 21:

Obtain an integral (or anti – derivative) of the $$\frac{u + 2}{\sqrt{4u ^{2} + 2 u + 3}}$$

Answer 21: Based on formulae given in Integrals

$$\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} \int \frac{2 (u + 2)}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \frac{1}{2} \int \frac{2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ = \frac{1}{2} \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du + \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ \int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} I_{1} + I_{2} \\$$ $$I_{1} = \int \frac{2 u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ Suppose,\; u ^{2} + 2 u + 3 = z \\ (2 u + 2) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 2 u + 3} ….. (2) \\ I_{2} = \int \frac{1}{\sqrt{u ^{2} + 2 u + 3}} \;du \\ u ^{2} + 2 u + 3 = u ^{2} + 2 u + 1 + 2 = (u + 1) ^{2} + (\sqrt{2}) ^{2} \\ I_{2} = \int \frac{1}{\sqrt{(u + 1) ^{2} + (\sqrt{2}) ^{2}}} \;du = log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | ….. (3)$$

Substituting equations (2) and (3) in equation (1), we get,

$$\int \frac{u + 2}{\sqrt{u ^{2} + 2 u + 3}} \;du = \frac{1}{2} (2 \sqrt{u ^{2} + 2 u + 3}) + log\; \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C \\ = \sqrt{u ^{2} + 2 u + 3} + log \left | (u + 1) + \sqrt{u ^{2} + 2 u + 3} \right | + C$$

Question 22:

Obtain an integral (or anti – derivative) of the $$\frac{u + 2}{\sqrt{u ^{2} – 2 u – 5}}$$

Answer 22: Based on formulae given in Integrals

$$Suppose,\; (u + 3) = A \frac{d}{du} (u ^{2} – 2u – 5) \\ (u + 3) = A (2 u – 2) + B \\$$

Equate the coefficients of u and the constants on both the sides, we get,

$$2 A = 1 => A = \frac{1}{2} \\ – 2 A + B = 3 => B = 4 \\ (u + 3) = \frac{1}{2} (2 u – 2) + 4 \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \int \frac{\frac{1}{2} (2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ = \frac{1}{2} \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du + 4 \int \frac{1}{u ^{2} – 2 u – 5} du \\$$ $$Suppose,\; I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \;and\; I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ \int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} I_{1} + 4 I_{2} …. (1) \\ Then, I_{1} = \int \frac{(2 u – 2) + 4}{u ^{2} – 2 u – 5} \;du \\ Suppose,\; u ^{2} – 2 u – 5 = Z \\ (2 u – 2) \;du = dz \\ I_{1} = \int \frac{dz}{z} \\ = log \left | z \right | + C \\ = log \left | u ^{2} – 2 u – 5 \right | + C …. (2)$$ $$I_{2} = \int \frac{1}{u ^{2} – 2 u – 5} du \\ = \int \frac{1}{(u ^{2} – 2 u + 1) – 6} du \\ = \int \frac{1}{(u – 1) ^{2} + (\sqrt{6}) ^{2}} \;du \\ = \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) …… (3)$$

Using equations (2) and (3) in equation (1), we get,

$$\int \frac{u + 3}{\sqrt{u ^{2} – 2 u – 5}} \;du = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + 4 \left [ \frac{1}{2 \sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right ) \right ] + C \\ = \frac{1}{2} log \left | u ^{2} – 2 u – 5 \right | + \frac{2}{\sqrt{6}} log \left ( \frac{u – 1 – \sqrt{6}}{u – 1 + \sqrt{6}} \right )$$

Question 23:

Obtain an integral (or anti – derivative) of the $$\frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}}$$

Answer 23: Based on formulae given in Integrals

$$Suppose,\; 5u + 3 = A \frac{d}{du} (u ^{2} + 4 u + 10) + B \\ 5u + 3 = A (2 u + 4) + B$$

Equate the coefficients of u and the constants on both the sides, we get,

$$2 A = 5 => A = \frac{5}{2} \\ 4 A + B = 3 => B = – 7 \\ 5 u + 3 = \frac{5}{2} (2 u + 4) – 7 \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{\frac{5}{2} ((2 u + 4) – 7)}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \frac{5}{2} \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du – 7 \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \;and\; I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ \int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} I_{1} – 7 I_{2} \;\; ….. (1)$$ $$I_{1} = \int \frac{2 u + 4}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ Suppose,\; u ^{2} + 4 u + 10 = z \\ (2 u + 4) \;du = dz \\ I_{1} = \int \frac{dz}{\sqrt{z}} = 2 \sqrt{z} = 2 \sqrt{u ^{2} + 4 u + 10} ….. (2) I_{2} = \int \frac{1}{\sqrt{u ^{2} + 4 u + 10}} \;du \\ = \int \frac{1}{\sqrt{(u ^{2} + 4 u + 4) + 6}} \;du \\ = \int \frac{1}{(u + 2) ^{2} + (\sqrt{6}) ^{2}} \;du \\ =log \left |(u + 2)\sqrt{u ^{2} + 4 u + 10} \right | ….. (3)$$

Substituting equations (2) and (3) in equation (1), we get,

$$\int \frac{5 u + 3}{\sqrt{u ^{2} + 4 u + 10}} \;du = \frac{5}{2} [2 \sqrt{u ^{2} + 4 u + 10}] – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\ = 5\; \sqrt{u ^{2} + 4 u + 10} – 7 log \left | (u + 2) + \sqrt{u ^{2} + 4 u + 10} \right | + C \\$$

Question 24: Which of the following below is the answer for $$\int \frac{du}{u ^{2} + 2 u + 2} \;du$$

$$(a) u tan ^{- 1} (u + 1) + C \\ (b) tan ^{- 1} (u + 1) + C \\ (c) (u + 1) tan ^{- 1} (u) + C \\ (d) tan ^{- 1} (u) + C$$

Answer 24: Based on formulae given in Integrals

$$\int \frac{du}{u ^{2} + 2 u + 2} \;du = \int \frac{du}{(u ^{2} + 2 u + 1) + 1} \\ = \int \frac{1}{(u + 1) ^{2} + (1) ^{2}} \;du \\ = \left [ tan ^{- 1} (u + 1) \right ] + C$$

Thus, (b) is the correct answer.

Question 25: Which of the following below is the answer for $$\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du$$

$$(a) \frac{1}{9} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (b) \frac{1}{2} sin ^{- 1} \frac{8 u – 9}{9} + C \\ (c) \frac{1}{3} sin ^{- 1} \frac{9 u – 8}{8} + C \\ (d) \frac{1}{2} sin ^{- 1} \frac{9 u – 8}{8} + C$$

Answer 25: Based on formulae given in Integrals

$$\int \frac{du}{\sqrt{9 u – 4 u ^{2}}} \;du \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u \right )}} \\ = \int \frac{du}{\sqrt{- 4 \left ( u ^{2} – \frac{9}{4} u + \frac{81}{64} – \frac{81}{64} \right )}} \\ = \int \frac{1}{\sqrt{- 4} \left [ \left ( u – \frac{9}{8} \right ) ^{2} – (\frac{9}{8}) ^{2} \right ]} \;du \\ = \frac{1}{2} \int \frac{1}{\sqrt{(\frac{9}{8}) ^{2} – \left ( u – \frac{9}{8} \right ) ^{2}}} \;du \\ = \frac{1}{2} \left [ sin ^{- 1} \left ( \frac{u – \frac{9}{8}}{\frac{9}{8}} \right ) \right ] + C \\ = \frac{1}{2} sin ^{- 1} \left ( \frac{8 u – 9}{9} \right ) + C$$

Thus, (b) is the correct answer.

Exercise 7.4

Question 1:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u + 1) (u + 2)}$$

Answer 1: Based on formulae given in Integrals

Suppose, $$\frac{u}{(u + 1) (u + 2)} = \frac{A}{u + 1} + \frac{B}{u + 2} \\ => u = A (u + 2) + B (u + 1)$$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 1

2 A + B = 0

On solving, we get,

A = – 1 and B = 2

$$\frac{u}{(u + 1) (u + 2)} = \frac{- 1}{u + 1} + \frac{2}{u + 2} \\ => \int \frac{u}{(u + 1) (u + 2)} \;du = \frac{- 1}{u + 1} + \frac{2}{u + 2} \;du \\ = – log \left | u + 1 \right | + 2 log \left | u + 2 \right | + C \\ = log \left ( u + 2 \right ) ^{2} – log \left | u + 1 \right | + C \\ = log \frac{(u + 1) ^{2}}{(u + 1)} + C$$

Question 2:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u ^{2} – 9}$$

Answer 2: Based on formulae given in Integrals

Suppose, $$\frac{1}{(u + 3) (u – 3)} = \frac{A}{u + 3} + \frac{B}{u – 3} \\ 1 = A (u – 3) + B (u + 3)$$

Equate the coefficients of u and the constants on both the sides, we get,

A + B = 0

– 3 A + 3 B = 1

On solving, we get

$$A = – \frac{1}{6} \; and\; B = \frac{1}{6} \\ \frac{1}{(u + 3) (u – 3)} = \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \\ => \int \frac{1}{u ^{2} – 9} \;du = \int \left ( \frac{- 1}{6 (u + 3)} + \frac{1}{6 (u – 3)} \right ) \;du \\ = – \frac{1}{6} log \left | u + 3 \right | + \frac{1}{6} log \left | u – 3 \right | + C \\ = \frac{1}{6} log \left | \frac{(u – 3)}{(u + 3)} \right | + C$$

Question 3:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)}$$

Answer 3: Based on formulae given in Integrals

Suppose, $$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ 3 u – 1 = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = – 1

On solving, we get,

A = 1, B = – 5, and C = 4

$$\frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} = \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \\ \int \frac{3 u – 1}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{(u – 1)} – \frac{5}{(u – 2} + \frac{4}{(u – 3)} \right \} \;du \\ = log \left | u – 1 \right | – 5 log \left | u – 2 \right | + 4 log \left | u – 3 \right | + C$$

Question 4:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u – 1) (u – 2) (u – 3)}$$

Answer 4: Based on formulae given in Integrals

Suppose, $$\frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{A}{(u – 1)} + \frac{B}{(u – 2} + \frac{C}{(u – 3)} \\ u = A (u – 2) (u – 3) + B (u – 1) (u – 3) + C (u – 1) (u – 2) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– 5 A – 4 B – 3 C = 1

6 A + 3 B + 2 C = 0

On solving, we get,

$$A = \frac{1}{2}, \;B = – 2 \;and \;C = \frac{3}{2} \\ \frac{u}{(u – 1) (u – 2) (u – 3)} = \frac{1}{2 (u – 1)} – \frac{2}{(u – 2} + \frac{3}{2 (u – 3)} \\ \int \frac{u}{(u – 1) (u – 2) (u – 3)} \;du = \int \left \{ \frac{1}{2 \;(u – 1)} – \frac{2}{(u – 2)} + \frac{3}{2 (u – 3)} \right \} \;du \\ = \frac{1}{2}\; log \left | u – 1 \right | – 2 log \left | u – 2 \right | + \frac{3}{2} log \left | u – 3 \right | + C$$

Question 5:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2 u}{u ^{2} + 3 u + 2}$$

Suppose, $$\frac{2 u}{u ^{2} + 3 u + 2} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} \\ 2 u = A (u + 2) + B (u + 1) …. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

2 A + B = 0

On solving, we get,

A = – 2 and B = 4

$$\frac{2 u}{(u + 1) (u + 2)} = \frac{- 2}{(u + 1)} + \frac{4}{(u + 1)} \\ \int \frac{2 u}{(u + 1) (u + 2)} \;du = \int \left \{ \frac{4}{(u + 1)} – \frac{2}{(u + 1)}\right \} \;du \\ = 4 log \left | u + 2 \right | – 2 log \left | u + 1 \right | + C$$

Question 6:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1 – u ^{2}}{u (1 – 2 u)}$$

Answer 6: Based on formulae given in Integrals

$$\frac{1 – u ^{2}}{u (1 – 2 u)}$$ is not a proper fraction.

Dividing (1 – u 2) by u (1 – 2 u), we get,

$$\frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left ( \frac{2 – u}{u (1 – 2 u)} \right ) \\ Suppose,\; \frac{2 – u}{u (1 – 2 u)} = \frac{A}{u} + \frac{B}{(1 -2 u)} \\ (2 – u) = A (1 – 2 u) + B u ……. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2 A + B = – 1

And A = 2

On solving, we get,

A = 2 and B = 3

$$\frac{2 – u}{u (1 – 2 u)} = \frac{2}{u} + \frac{3}{(1 -2 u)} \\ Using\; in\; equation\; (1), we\; get, \\ \frac{1 – u ^{2}}{u (1 – 2 u)} = \frac{1}{2} + \frac{1}{2} \left \{ \frac{2}{u} + \frac{3}{(1 -2 u)} \right \} \\ = \frac{u}{2} + log \left | u \right | + \frac{3}{2 (- 2)}\; log \left | 1 – 2u \right | + C \\ = \frac{u}{2} + log \left | u \right | – \frac{3}{4}\; log \left | 1 – 2u \right | + C$$

Question 7:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u ^{2} + 1) (u – 1)}$$

Answer 7: Based on formulae given in Integrals

Suppose, $$\frac{u}{(u ^{2} + 1) (u – 1)} = \frac{A u + B}{(u ^{2} + 1)} + \frac{C}{u – 1} \\ u = (A u + B) (u – 1) + C (u^{2} + 1) \\ u = A u ^{2} – A u + B u – B + C u^{2} + C$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

– A + B = 1

– B + C = 0

On solving, we get,

$$A = – \frac{1}{2},\; B = \frac{1}{2}, and\; C = \frac{1}{2} \\ Using\; equation\; (1),\; we\; get\; \\ \frac{u}{(u ^{2} + 1) (u – 1)} = \frac{\left ( – \frac{1}{2} u + \frac{1}{2} \right )}{(u ^{2} + 1)} + \frac{\frac{1}{2}}{(u – 1)} \\$$ $$\int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{2} \int \frac{u}{(u ^{2} + 1)} \;du + \frac{1}{2} \int \frac{1}{(u ^{2} + 1)} + \frac{1}{2} \int \frac{1}{2} \int \frac{1}{u – 1} \;du \\ = – \frac{1}{4} \int \frac{2 u}{(u ^{2} + 1)} + \frac{1}{2} tan ^{- 1} u + \frac{1}{2} log \left | u – 1 \right | + C \\ \int \frac{2 u}{(u ^{2} + 1)} \;du, let\; (u ^{2} + 1) = z => 2u\; du = dz \\ \int \frac{2 u}{(u ^{2} + 1)} \;du = \int \frac{dz}{z} = log \left | z \right | = log\; \left | u^{2} + 1 \right | \\ \int \frac{u}{(u ^{2} + 1) (u – 1)} \;du = – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + \frac{1}{2} log \left | u – 1 \right | + C \\ = \frac{1}{2} log \left | u – 1 \right | – \frac{1}{4} log \left | (u ^{2} + 1) \right | + \frac{1}{2} tan ^{- 1}u + C$$

Question 8:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u}{(u – 1)^{2} (u + 2)}$$

Answer 8: Based on formulae given in Integrals

$$\frac{u}{(u – 1) c^{2} (u + 2)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1) ^{2}} + \frac{C}{u + 2} \\ u = A (u – 1) (u + 2) + B (u + 2) + C (u – 1)^{2} \\$$

Putting u = 1, we get,

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

A + B – 2 C = 1

– 2 A + 2 B + C = 0

On solving, we get,

$$A = \frac{2}{9},\; B = \frac{1}{3} \;and\; C = – \frac{2}{9}$$ $$\frac{u}{(u – 1)^{2} (u + 2)} = \frac{2}{9 \;(u – 1)} + \frac{1}{3 \;(u – 1) ^{2}} – \frac{2}{9 (u + 2)} \\ \int \frac{u}{(u – 1)^{2} (u + 2)} \;du = \frac{2}{9} \int \frac{1}{(u – 1)} \;du + \frac{1}{3} \int \frac{1}{(u – 1) ^{2}} \;du – \frac{2}{9} \int \frac{1}{(u + 2)} \;du \\ = \frac{2}{9} log\; \left | u – 1 \right | + \frac{1}{3} \left ( \frac{- 1}{u – 1} \right ) – \frac{2}{9} log \left | u + 2 \right | + C \\ = \frac{2}{9} log\; \left | \frac{u – 1}{u + 2} \right | – \frac{1}{3 (u – 1)} + C$$

Question 9:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3 u + 5}{u^{3} – u^{2} – u + 1}$$

Answer 9: Based on formulae given in Integrals

$$\frac{3 u + 5}{u^{3} – u^{2} – u + 1} = \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \\ Suppose,\; \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{A}{(u – 1)} + \frac{B}{(u – 1)^{2}} + \frac{C}{(u + 1)} \\ 3 u + 5 = A (u – 1) (u + 1) + B (u + 1) + C (u – 1) ^{2} \\ 3 u + 5 = A (u^{2} – 1) + B (u + 1) + C (u^{2} + 1 – 2 u) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + C = 0

B – 2 C = 3

– A + B + C = 5

On solving, we get,

B = 4

$$A = – \frac{1}{2} \;and\; C = \frac{1}{2} \\ \frac{3 u + 5}{(u – 1)^{2} (u + 1)} = \frac{- 1}{2 (u – 1)} + \frac{4}{(u – 1)^{2}} + \frac{1}{2 (u + 1)} \\ \int \frac{3 u + 5}{(u – 1)^{2} (u + 1)} \;du = – \frac{1}{2} \int \frac{1}{(u – 1)} \;du + 4 \int \frac{1}{(u – 1)^{2}} \;du + \frac{1}{2} \int \frac{1}{(u + 1)} \;du \\ = – \frac{1}{2} log\; \left | u – 1 \right | + 4 \left ( \frac{- 1}{u – 1} \right ) + \frac{1}{2} log \left | u + 1 \right | + C \\ = \frac{1}{2} log \left | \frac{u + 1}{u – 1} \right | – \frac{4}{(u – 1)} + C$$

Question 10:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)}$$

Answer 10: Based on formulae given in Integrals

$$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)}$$

Suppose, $$\frac{2 u – 3}{(u^{2} – 1) (2u + 3)} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{C}{(2u + 3)} \\ (2u – 3) = A\; (u – 1)(2 u + 3) + B\; (u + 1) (2 u + 3) + C\; (u + 1) (u – 1) \\ (2u – 3) = A\; (2 u^{2} + u – 3) + B\; (2 u^{2} + 5u + 3) + C\; (u^{2} – 1) \\ (2u – 3) = (2A + 2B + C) u^{2} + (A + 5B)u + (- 3A + 3B – C)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

2A + 2B + C = 1

A + 5B = 2

– 3A + 3B – C = – 3

On solving, we get,

$$\frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2 (u + 1)} – \frac{1}{10 (u – 1)} – \frac{24}{5 (2u + 3)} \\ \frac{2 u – 3}{(u + 1)(u – 1)(2u + 3)} = \frac{5}{2} \int \frac{1}{(u + 1)} \;du – \frac{1}{10} \int \frac{1}{ (u – 1)} \;du – \frac{24}{5} \int \frac{1}{ (2u + 3)} \;du \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{24}{5 \times 2} log\; \left | 2u + 3 \right | + C \\ = \frac{5}{2} log\; \left | u + 1 \right | – \frac{1}{10} log\; \left | u – 1 \right | – \frac{12}{5} log\; \left | 2u + 3 \right | + C$$

Question 11:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{5 u}{(u + 1) (u^{2} – 4)}$$

Answer 11: Based on formulae given in Integrals

$$\frac{5 u}{(u + 1) (u^{2} – 4)} = \frac{5 u}{(u + 1)(u + 2)(u – 2)} \\ Suppose,\; \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{A}{(u + 1)} + \frac{B}{(u + 2)} + \frac{C}{(u – 2)} \\ 5u = A (u + 2)(u – 2) + B (u + 1)(u – 2) + C (u + 1)(u + 2) …. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– B + 3C = 5 and

– 4A – 2B + 2C = 0

On solving, we get,

$$A = \frac{5}{3}, B = – \frac{5}{2} \;and\; C = \frac{5}{6} \\ \frac{5 u}{(u + 1)(u + 2)(u – 2)} = \frac{5}{3 (u + 1)} – \frac{5}{2 (u + 2)} + \frac{5}{6 (u – 2)} \\ \int \frac{5 u}{(u + 1)(u + 2)(u – 2)} \;du = \frac{5}{3} \frac{1}{(u + 1)} \;du – \frac{5}{2} \frac{1}{(u + 2)} \;du + \frac{5}{6} \frac{1}{(u – 2)} \;du \\ = \frac{5}{3} log\; \left | u + 1 \right | – \frac{5}{2} log\; \left | u + 2 \right | + \frac{5}{6} log\; \left | u – 2 \right | + C$$

Question 12:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{u^{3} + u + 1}{u^{2} – 1}$$

Answer 12: Based on formulae given in Integrals

$$\frac{u^{3} + u + 1}{u^{2} – 1}$$ is not a proper fraction.

So, dividing (u3 + u + 1) by u2 – 1, we get,

$$\frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{2u + 1}{u^{2} – 2} \\ Suppose,\; \frac{2u + 1}{u^{2} – 2} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} \\ 2u + 1 = A (u – 1) + B (u + 1) ….. (1)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 2

– A + B = 1

On solving, we get,

$$A = \frac{1}{2} \;and\; B = \frac{3}{2} \\ \frac{u^{2} + u + 1}{u^{2} – 1} = u + \frac{1}{2 (u + 1)} + \frac{3}{2 (u – 1)} \\ Integrating\; on\; both\; the\; sides,\; we\; get, \int \frac{u^{2} + u + 1}{u^{2} – 1} \;du = \int u \;du + \frac{1}{2} \int \frac{1}{u + 1} \;du + \frac{3}{2} \frac{1}{(u – 1)} \;du \\ = \frac{u ^{2}}{2} + \frac{1}{2} log\; \left | u + 1 \right | – \frac{3}{2} log\; \left | u – 1 \right | + C$$

Question 13:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2}{(1 – u)(1 + u ^{2})}$$

Answer 13: Based on formulae given in Integrals

Suppose, $$\frac{2}{(1 – u)(1 + u ^{2})} = \frac{A}{1 – u} + \frac{Bu + C}{1 + u ^{2}} \\ 2 = A (1 + u ^{2}) + (Bu + C)(1 – u) \\ 2 = A + A\; u^{2} + Bu – B\; u^{2} + C – Cu$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A – B = 0

B – C = 0

A + C = 2

On solving, we get,

A = 1, B = 1 and C = 1

$$\frac{2}{(1 – u)(1 + u ^{2})} = \frac{1}{1 – u} + \frac{u + 1}{1 + u ^{2}} \\ \int \frac{2}{(1 – u)(1 + u ^{2})} \;du = \int \frac{1}{1 – u} \;du + \int \int \frac{u}{1 + u ^{2}} \;du + \int \frac{1}{1 + u ^{2}} \;du \\ = – log\; \left | u – 1 \right | + \frac{1}{2} log\; \left | 1 + u^{2} \right | + tan ^{- 1} \;u + C$$

Question 14:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{3u – 1}{(u + 2)^{2}}$$

Answer 14: Based on formulae given in Integrals

Suppose, $$\frac{3u – 1}{(u + 2)^{2}} = \frac{A}{(u + 2)} + \frac{B}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 3

2A + B = – 1

B = – 7

$$\frac{3u – 1}{(u + 2)^{2}} = \frac{3}{(u + 2)} – \frac{7}{(u + 2)^{2}} \\ 3u – 1 = A (u + 2) + B \frac{3u – 1}{(u + 2)^{2}} = 3 \int \frac{1}{(u + 2)} \;du – 7 \int \frac{u}{(u + 2)^{2}} \;du \\ = 3\; log\; \left | u + 2 \right | – 7 \left ( \frac{- 1}{(u\; + \;2)} \right ) + C \\ = 3\; log\; \left | u + 2 \right | + \left ( \frac{7}{(u\; + \;2)} \right ) + C$$

Question 15:

Getting integral (or anti – derivative) of the following rational number $$\frac{1}{u^{4} – 1}$$

$$\frac{1}{u^{4} – 1} = \frac{1}{(u^{2} – 1)(u^{4} + 1)} = \frac{1}{(u + 1) (u – 1)(1 + u^{2})} \\ Suppose,\; \frac{1}{(u + 1) (u – 1)(1 + u^{2})} = \frac{A}{(u + 1)} + \frac{B}{(u – 1)} + \frac{Cu + D}{(u^{2} + 1)} \\ 1 = A (u + 1)(u^{2} + 1) + B (u + 1)(u^{2} + 1) + (Cu + D) (u^{2} – 1) \\ 1 = A (u^{3} + u – u^{2} – 1) + B (u^{3} + u + u^{2} + 1) + Cu^{2} + Du^{2} – Cu – D \\ 1 = (A + B + C) u^{3} + (- A + B + D) u^{2} + (A + B – C)u + (- A + B – D)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B + C = 0

– A + B + D = 0

A + B – C = 0

– A + B – D = 1

On solving, we get,

$$A = – \frac{1}{4}, B = \frac{1}{4}, C = 0 \;and\; D = – \frac{1}{2} \\ \frac{1}{u^{4} – 1} = \frac{- 1}{4 (u + 1)} + \frac{1}{4 (u – 1)} – \frac{1}{2 (u^{2} + 1)} \\ \int \frac{1}{u^{4} – 1} \;du = – \frac{1}{4} \int \frac{1}{(u + 1)} \;du + \frac{1}{4} \int \frac{1}{(u – 1)} \;du + \frac{1}{2} \int \frac{1}{(u^{2} + 1)} \\ = – \frac{1}{4} log\; \left | u + 1 \right | + \frac{1}{4} log\; \left | u – 1 \right | – \frac{1}{2} tan^{- 1} \;u + C \\ = \frac{1}{4} log\; \left | \frac{ u – 1}{u + 1} \right | – \frac{1}{2} tan^{- 1} \;u + C$$

Question 16:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u (u^{m} + 1)}$$

[Hint: multiply denominator and numerator by u n – 1 and put un = z]

Answer 16: Based on formulae given in Integrals

$$\frac{1}{u (u^{m} + 1)}$$

Multiplying denominator and numerator by u n – 1, we get,

$$\frac{1}{u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m – 1}. u (u^{m} + 1)} = \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \\ Suppose,\; u^{m} = z => u ^{m – 1} \;du = dz \\ \int \frac{1}{u (u^{m} + 1)} \;du = \int \frac{u ^{m – 1}}{u ^{m} (u^{m} + 1)} \;du = \frac{1}{m} \int \frac{1}{z (z + z)} \;du \\ Suppose,\; \frac{1}{z (z + z)} = \frac{A}{z} + \frac{B}{(z + 1)} \\ 1 = A (1 + z) + Bz \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = 1 and B = – 1

$$\frac{1}{z (z + z)} = \frac{1}{z} – \frac{1}{(z + 1)} \\ \int \frac{1}{u (u^{m} + 1)} \;du = \frac{1}{m} \int \left \{ \frac{1}{z} – \frac{1}{(z + 1)} \right \} + C \\ = \frac{1}{m} \left [ log\; \left | u^{m} \right | – log\; \left | u^{m} + 1 \right | \right ] + C \\ = \frac{1}{m} log\; \left | \frac{u^{m}}{u^{m} + 1} \right |$$

Question 17:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$$

[Hint: Put sin u = z]

Answer 17: Based on formulae given in Integrals

$$\frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)}$$

Suppose, sin u = z => cos u du = dz

$$\int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \frac{dz}{(1 – z)(2 – z)} \\ Suppose,\; \frac{1}{(1 – z)(2 – z)} = \frac{A}{(1 – z)} + \frac{B}{(2 – z)} \\ 1 = A (2 – z) + B (1 – z)$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

– 2A – B = 0

2A + B = 1

On solving, we get,

A = 1 and B = – 1

$$\frac{1}{(1 – z)(2 – z)} = \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \\ \int \frac{cos\; u}{(1 – sin\; u)(2 – sin\; u)} \;du = \int \left \{ \frac{1}{(1 – z)} – \frac{1}{(2 – z)} \right \} \;dz \\ = – log\; \left | 1 – z \right | + log\; \left | 2 – z \right | + C \\ = log\; \left | \frac{2 – z}{1 – z} \right | + C \\ = log\; \left | \frac{2 – sin\; u}{1 – sin\; u} \right | + C$$

Question 18:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)}$$

Answer 18: Based on formulae given in Integrals

$$\frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} \\ Suppose,\; \frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{Au + B}{(u^{2} + 3)} + \frac{Cu + D}{(u^{2} + 4)} \\ (4 u^{2} + 10) = (Au + B)(u^{2} + 4) + (Cu + D)(u^{2} + 3) \\ (4 u^{2} + 10) = A u^{3} + 4 Au + B u^{2} + 4 B + C u^{3} + 3 Cu + D u^{2} + 3D \\ (4 u^{2} + 10) = (A + C) u^{3} + (B + D) u^{2} + (4A + 3C) u + (4B + 3D)$$

Equate the coefficients of u3, u2, u and the constants on both the sides, we get,

A + C = 0

B + D = 4

4 A + 3 C = 0

4 B + 3 D = 10

On solving, we get,

A = 0, B = – 2, C = 0 and D = 6

$$\frac{(4 u^{2} + 10)}{(u^{2} + 3)(u^{2} + 4)} = \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \\ \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} = 1 – \left ( \frac{- 2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right ) \\ \int \frac{(u^{2} + 1)(u^{2} + 2)}{(u^{2} + 3)(u^{2} + 4)} \;du = \int \left \{ 1 + \frac{2}{(u^{2} + 3)} + \frac{6}{(u^{2} + 4)} \right \} \\ = \int \left \{ 1 + \frac{2}{u ^{2} + (\sqrt{3})^{2}} – \frac{6}{u^{2} + 2^{2}} \right \} \;du \\ = u + 2 \left ( \frac{1}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} \right ) – 6 \left ( \frac{1}{2} \;tan ^{- 1} \frac{u}{2} \right ) + C \\ = u + \frac{2}{\sqrt{3}} \;tan ^{- 1} \frac{u}{\sqrt{3}} – 3 \;tan ^{- 1} \frac{u}{2} + C$$

Question 19:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$$

Answer 19: Based on formulae given in Integrals

$$\frac{2\; u}{(u^{2} + 1)(u^{2} + 3)}$$

Suppose, u2 = z

2u du = dz

$$\int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \frac{dz}{(z + 1)(z + 3)} …. (1) \\ Suppose,\; \frac{1}{(z + 1)(z + 3)} = \frac{A}{(z + 1)} + \frac{B}{(z + 3)} \\ 1 = A\; (z + 3) + B\; (z + 1) …. (1) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A + B = 0

3 A + B = 1

On solving, we get,

$$A = \frac{1}{2} \;and\; B = – \frac{1}{2} \\ \frac{1}{(z + 1)(z + 3)} = \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \\ \int \frac{2\; u}{(u^{2} + 1)(u^{2} + 3)} \;du = \int \left \{ \frac{1}{2\; (z + 1)} – \frac{1}{2\; (z + 3)} \right \} \;dz \\ = \frac{1}{2} log \left | (z + 1) \right | – \frac{1}{2} log \left | (z + 3) \right | + C \\ = \frac{1}{2} log \left | \frac{z + 1}{z + 3} \right | + C \\ = \frac{1}{2} log \left | \frac{u^{2} + 1}{u^{2} + 3} \right | + C$$

Question 20:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{u (u^{4} – 1)}$$

Answer 20: Based on formulae given in Integrals

$$\frac{1}{u (u^{4} – 1)}$$

Multiplying denominator and numerator by u3, we get,

$$\frac{1}{u (u^{4} – 1)} = \frac{u^{3}}{u^{4} (u^{4} – 1)} \\ \int \frac{1}{u (u^{4} – 1)} \;du = \int \frac{u^{3}}{u^{4} (u^{4} – 1)} \;du \\ Suppose,\; u^{4} = t => 4 u^{3} \;du = dz \\ \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \frac{dz}{z (z – 1)} \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z – 1} \\ 1 = A (z – 1) + Bz ….. (1) \\$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ = \int \frac{1}{u (u^{4} – 1)} \;du = \frac{1}{4} \int \left ( \frac{- 1}{z} + \frac{1}{z – 1} \right ) \;dz \\ = \frac{1}{4} \left [ – log\; \left | z \right | + log\; \left | z – 1 \right | \right ] + C \\ = \frac{1}{4} log\; \left | \frac{z – 1}{z} \right | + C \\ = \frac{1}{4} log\; \left | \frac{u^{4} – 1}{u^{4}} \right | + C$$

Question 21:

Obtain an integral (or anti – derivative) of the following rational number $$\frac{1}{(e^{u} – 1)}$$

Answer 21: Based on formulae given in Integrals

$$\frac{1}{(e^{u} – 1)}$$

Suppose, eu = z

eu du = dz

$$\int \frac{1}{(e^{u} – 1)} \;du = \int \frac{1}{z – 1} \times \frac{dz}{z} = \int \frac{1}{z (z – 1)} \;dz \\ Suppose,\; \frac{1}{z (z – 1)} = \frac{A}{z} + \frac{B}{z – 1} \\ 1 = A (z – 1) + Bz$$

Equate the coefficients of u2, u and the constants on both the sides, we get,

A = – 1 and B = 1

$$\frac{1}{z (z – 1)} = \frac{- 1}{z} + \frac{1}{z – 1} \\ \int \frac{1}{z (z – 1)} \;du = log\; \left | \frac{z – 1}{z} \right | + C \\ = log\; \left | \frac{e^{u} – 1}{e^{u}} \right | + C$$

Question 22: Which of the following below is an integral of $$\frac{u \;du}{(u – 1)(u – 2)}$$

$$(a) log\; \left | \frac{(u – 1)^{2}}{u – 2} \right | + C \\ (b) log\; \left | \frac{(u – 2)^{2}}{u – 2} \right | + C \\ (c) log\; \left | (\frac{(u – 1)}{u – 2}) ^{2} \right | + C \\ (d) log\; \left | (u – 1)(u – 2) \right | + C$$

Answer 22: Based on formulae given in Integrals

$$Suppose,\; \frac{u \;du}{(u – 1)(u – 2)} = \frac{A}{(u – 1)} + \frac{B}{u – 2} \\ u = A (u – 2) + B (u – 1) …. (1)$$

Equate the coefficients of u and the constants on both the sides, we get,

A = – 1 and B = 2

$$\frac{u \;du}{(u – 1)(u – 2)} = \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \\ \int \frac{u \;du}{(u – 1)(u – 2)} \;d= \left \{ \frac{- 1}{(u – 1)} + \frac{2}{u – 2} \right \} \;du \\ = – log\; \left | u – 1 \right | + 2\; log\; \left | u – 2 \right | + C \\ = log\; \left | \frac{(u – 2) ^{2}}{u – 1} \right | + C$$

Hence, option (b) is the correct answer.

Question 23: Which of the following below is an integral of $$\int \frac{du}{u (u^{2} + 1)} \;du$$

$$(a) log\; \left | u \right | – \frac{1}{2} log\; (u^{2} + 1) + C \\ (b) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (c) – log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\ (d) log\; \left | u \right | + \frac{1}{2} log\; (u^{2} + 1) + C \\$$

Answer 23: Based on formulae given in Integrals

$$Suppose,\; \frac{1}{u (u^{2} + 1)} = \frac{A}{u} + \frac{Bu + C}{u^{2} + 1} \\ 1 = A (u^{2} + 1) + (Bu + C) u$$

Equate the coefficients of u2, u and the constants on both the sides,

A + B = 0

C = 0

A = 1

Solving, we get,

A =1, B = – 1, and C = 0

$$\frac{1}{u (u^{2} + 1)} = \frac{1}{u} + \frac{- U}{u^{2} + 1} \\ \int \frac{1}{u (u^{2} + 1)} \;du = \int \left \{ \frac{1}{u} – \frac{u}{u^{2} + 1} \right \} \;du \\ = log\; \left | u \right | – \frac{1}{2} log\; \left | u^{2} + 1 \right | + C$$

Hence, option (a) is the correct answer.

Exercise 7.6

Question 1:

Integral of u sin u.

Answer 1: Based on formulae given in Integrals

Suppose, I = $$\int u\; sin\; u \;du$$

Integrating by parts by taking u as first function and sin u as second function, we get,

$$I = u\;\int sin\; u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; u \;du \right \} \;du \\ = u (- cos\; u) – \int 1. (- cos\; u) \;du = – u\; cos\; u + sin\; u + C$$

Question 2:

Obtain an integral of u sin 3u.

Answer 2: Based on formulae given in Integrals

Suppose, I = $$\int u\; sin\; u\; du$$

Integrating by parts by taking u as first function and sin 3u as second function, we get,

$$I = u\;\int sin\; 3u \;du – \int \left \{ \left ( \frac{d}{du} \;u \right ) \int sin\; 3u \;du \right \} \;du \\ = u (\frac{- cos\; 3u}{3}) – \int 1. (\frac{- cos\; 3u}{3}) \;du \\ = \frac{- u\; cos\; 3u}{3} + \frac{1}{9} sin\; 3u + C$$

Question 3:

Obtain an integral of $$u^{2}. e^{u}$$

Answer 3: Based on formulae given in Integrals

Suppose, I = $$\int u^{2}. e^{u} \;du$$

Integrating by parts by taking u2 as first function and eu as second function, we get,

$$I = u^{2} \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du} u^{2} \right ) \int e^{u} \;du \right \} \;du \\ = u^{2} e^{u} – \int 2u\; e^{u} \;du \\ = u^{2} e^{u} – 2 \int u\; e^{u} \;du \\ Integrating\; by\; parts\;, we\; get, \\ = u^{2} e^{u} – 2 \left [u\; \int e^{u} \;du – \int \left \{ \left ( \frac{d}{du}\; u \right ). \int e^{u} \;du \right \} \right ] \;du \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – \int e^{u} \;du \right ] \\ = u^{2} e^{u} – 2 \left [u\; e^{u} – e^{u} \right ] \\ = u^{2} e^{u} – 2 u\; e^{u} – 2 e^{u} + C \\ = e^{u} (u^{2} – 2u + 2) + C$$

Question 4:

Obtain an integral of u log u.

Answer 4: Based on formulae given in Integrals

Suppose, I = $$\int u\; log\; u \;du$$

Integrating by parts by taking log u as first function and u as second function, we get,

$$I = log\; u \int u\; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u \;du \right \} \;du \\ = log\; u. \frac{u^{2}}{2} – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; u}{2} = \frac{u^{2}}{4} + C$$

Question 5:

Obtain an integral of u log 2u.

Answer 5: Based on formulae given in Integrals

Suppose, I = $$\int u\; log\; 2u \;du$$

Integrating by parts by taking log 2 u as first function and u as second function, we get,

$$I = log\; 2u \int u\; du – \int \left \{ \left ( \frac{d}{du} 2 log\; u \right ) \int u \;du \right \} \;du \\ = log\; 2u. \frac{u^{2}}{2} – \int \frac{2}{2 u}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \int \frac{u}{2} \;du \\ = \frac{u^{2} log\; 2u}{2} – \frac{u^{2}}{4} + C$$

Question 6:

Obtain an integral of u2 log u.

Answer 6: Based on formulae given in Integrals

Suppose, I = $$\int u^{2}\; log\; u \;du$$

Integrating by parts by taking log u as first function and u 2 as second function, we get,

$$I = log\; u \int u ^{2} \; du – \int \left \{ \left ( \frac{d}{du} log\; u \right ) \int u^{2} \;du \right \} \;du \\ = log\; u. \frac{u^{3}}{3} – \int \frac{1}{u}. \frac{u^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{3} – \int \frac{u ^{2}}{3} \;du \\ = \frac{u^{3} log\; u}{2} – \frac{u^{3}}{9} + C$$

Question 7:

Obtain an integral of u sin – 1 u.

Answer 7: Based on formulae given in Integrals

Suppose, I = $$\int u\; sin ^{- 1}\; u \;du$$

Integrating by parts by taking sin – 1 u as first function and u as second function, we get,

$$I = sin ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = sin ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \frac{1 – u^{2}}{\sqrt{1 – u^{2}}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du$$ $$= \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} – \frac{1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \int \sqrt{1 – u^{2}} \;du – \int \frac{1}{\sqrt{1 – u^{2}}} \;du \right \} \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{1}{2} \left \{ \frac{u}{2} \sqrt{1 – u^{2}} + \frac{1}{2} sin ^{- 1} \;u – sin ^{- 1} \;u \right \} + C \\ = \frac{u^{2} sin ^{- 1}\; u}{2} + \frac{u}{4} \sqrt{1 – u^{2}} + \frac{1}{4} sin ^{- 1} \;u – \frac{1}{2} sin ^{- 1} \;u + C \\ = \frac{1}{4} (2u^{2} – 1) sin ^{- 1} \;u + \frac{u}{4} \sqrt{1 – u ^{2}} + C$$

Question 8:

Obtain an integral of u tan – 1 u

Answer 8: Based on formulae given in Integrals

Suppose, I = $$\int u \;tan ^{- 1}\; u\; du$$

Integrating by parts by taking tan – 1 u as first function and u as second function, we get,

$$I = tan ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} tan ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = tan ^{- 1}\; u \;\frac{u^{2}}{2} – \int \frac{1}{1 + u^{2}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{ u^{2}}{1 + u^{2}} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ \frac{u^{2} + 1}{1 + u^{2}} – \frac{1}{1 + u^{2}} \right \} \;du$$ $$= \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int \left \{ 1 – \frac{1}{1 + u^{2}} \right \} \;du \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{1}{2} \int (u – tan ^{- 1} \;u) + C \\ = \frac{u^{2} tan ^{- 1}\; u}{2} – \frac{u}{2} + \frac{1}{2}\; tan ^{- 1}\; u + C$$

Question 9:

Obtain an integral of u cos – 1 u

Answer 9: Based on formulae given in Integrals

Suppose, I = $$\int u\; cos ^{- 1}\; u\; du$$

Integrating by parts by taking cos – 1 u as first function and u as second function, we get,

$$I = cos ^{- 1}\; u \int u \;du – \int \left \{ \left ( \frac{d}{du} cos ^{- 1}\; u \right ) \int u \;du \right \} \;du \\ = cos ^{- 1}\; u \frac{u^{2}}{2} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. \frac{u^{2}}{2} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} \int \frac{1 – u^{2} + 1}{\sqrt{1 – u^{2}}} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} + \frac{1}{2} \int \left \{ \sqrt{1 – u^{2}} + \frac{- 1}{\sqrt{1 – u^{2}}} \right \} \;du \\ = \frac{u^{2} cos ^{- 1}\; u}{2} – \frac{1}{2} I_{1} – \frac{1}{2}\; cos ^{- 1} \;u ….. (1)$$ $$where\; I_{1} = \int \sqrt{1 – u^{2}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{d}{du}\; \sqrt{1 – u^{2}} \int u\; du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- 2u}{2 \sqrt{1 – u^{2}}}. u \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{- u^{2}}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \int \frac{1 – u^{2} – 1}{\sqrt{1 – u^{2}}} \;du \\ I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ \int \sqrt{1 – u^{2}} \;du + \int \frac{- du}{\sqrt{1 – u^{2}}} \right \}$$ $$I_{1} = u \;\sqrt{1 – u^{2}} – \left \{ I_{1} + cos ^{- 1} \;u \right \} \\ 2 I_{1} = u \;\sqrt{1 – u^{2}} – cos ^{- 1} \;u \\ I_{1} = \frac{u}{2} \;\sqrt{1 – u^{2}} – \frac{1}{2} cos ^{- 1} \;u \\ Using\; in\; equation\; (1),\; we\; get, \\ I = \frac{u^{2} sin ^{- 1}\; u}{2} – \frac{1}{2} \left ( \frac{u}{2}\; \sqrt{1 – u^{2}} – \frac{1}{2}\; cos ^{- 1} \;u \right ) – \frac{1}{2}\; cos ^{- 1} \;u \\ = \frac{(2 u^{2} – 1)}{4}\; cos ^{- 1} \;u – \frac{u}{4}\; \sqrt{1 – u^{2}} + C$$

Question 10:

Obtain an integral of (sin – 1 u) 2

Answer 10: Based on formulae given in Integrals

Suppose, I = $$\int (sin ^{- 1} \;u) ^{2} . 1\; du$$

Integrating by parts by taking (sin – 1 u) 2 as first function and 1 as second function, we get,

$$sin ^{- 1} \;u \int 1 \;du – \int \left \{ \frac{d}{du} (sin ^{- 1} \;u)^{2}. \int 1. du \right \} \;du \\ = (sin ^{- 1} \;u)^{2}. \;u – \int \frac{2\; (sin ^{- 1} \;u)}{\sqrt{1 – u^{2}}}. \;u \;du \\ = u. (sin ^{- 1} \;u)^{2} + \int sin ^{- 1} \;u. \left ( \frac{- 2u}{\sqrt{1 – u^{2}}} \right ) \;du \\ = u. (sin ^{- 1} \;u)^{2} + \left [sin ^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du} sin ^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du\right \} \;du \right ] \\ = u. (sin ^{- 1} \;u)^{2} + \left [ sin ^{- 1} \;u. 2\sqrt{1 – u^{2}} – \int \frac{1}{\sqrt{1 – u^{2}}} . 2 \sqrt{1 – u^{2}} \;du \right ]$$ $$= u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u – \int 2 \;du \\ = u. (sin ^{- 1} \;u)^{2} + 2 \sqrt{1 – u^{2}}\; sin^{- 1} \;u -2u + C$$

Question 11:

Obtain an integral of $$\frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}}$$

Answer 11: Based on formulae given in Integrals

Suppose, I = $$\int \frac{u\; cos^{- 1} \;u}{\sqrt{1 – u^{2}}} \;du \\ I = \frac{- 1}{2} \int \frac{- 2u}{\sqrt{1 – u^{2}}}.\; cos ^{- 1} u \;du$$

Integrating by parts by taking cos – 1 u as first function and $$\frac{- 2u}{\sqrt{1 – u^{2}}}$$ as second function, we get,

$$I = \frac{- 1}{2} \left [cos^{- 1} \;u \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du – \int \left \{ \left ( \frac{d}{du}\; cos^{- 1} \;u \right ) \int \frac{- 2u}{\sqrt{1 – u^{2}}} \;du \right \} \;du \right ] \\ = \frac{- 1}{2} \left [ cos^{- 1} \;u. 2\; \sqrt{1 – u^{2}} – \int \frac{- 1}{\sqrt{1 – u^{2}}}. 2\; \sqrt{1 – u^{2}} \;du \right ] \\ = \frac{- 1}{2} \left [2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + \int 2 \;du \right ] \\ = \frac{- 1}{2} \left [ 2\; \sqrt{1 – u^{2}}\; cos^{- 1} \;u + 2u \right ] + C \\ = – \left [\sqrt{1 – u^{2}} \;cos^{- 1} \;u + u \right ] + C$$

Question 12:

Obtain an integral of u sec 2 u

Answer 12: Based on formulae given in Integrals

Suppose, I = $$\int u\; sec^{2} \;u \;du$$

Integrating by parts by taking u as first function and sec 2 u as second function, we get,

$$u \int sec^{2} \;u \;du – \int \left \{ \left \{ \frac{d}{du}. u \right \} \int sec^{2} \;u \;du \right \} \;du \\ = u\; tan\; u – \int 1. \;tan \;u \;du \\ = u\; tan\; u – log\; \left | cos\; u \right | + C$$

Question 13:

Obtain an integral of tan – 1 u

Answer 13: Based on formulae given in Integrals

Suppose, I = $$\int tan ^{- 1} \;u \;du$$

Integrating by parts by taking tan – 1 u as first function and 1 as second function, we get,

$$tan ^{- 1} \;u \int 1 \;du – \int \left \{ \left ( \frac{d}{du} \;tan ^{- 1} \;u \right ) \int 1 \;du \right \} \;du \\ = tan ^{- 1} \;u . \;u – \int \frac{1}{1 + u^{2}}. \;u \;du \\ = tan ^{- 1} \;u . \;u – \frac{1}{2} \int \frac{2u}{1 + u^{2}}. \;du \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; \left | 1 + u^{2} \right | + C \\ = u\; tan ^{- 1} \;u – \frac{1}{2} \;log\; (1 + u^{2}) + C$$

Question 14:

Obtain an integral of u (log u) 2.

Answer 14: Based on formulae given in Integrals

Suppose in this case, I =

Integrating the equation by parts by taking (log u) 2 as first function and 1 as second function,

$$I = (log \;u) ^{2} \int u \;du – \int \left [ \left \{ (\frac{d}{du} \; log \;u) ^{2} \right \} \int u\; du \right ] \;du \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [ \int 2 log\; u. \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \int u\; log\; u \;du$$

Integrating by parts, we get,

$$I = \frac{u^{2}}{2} (log \;u) ^{2} \int u \;du – \left [log \;u \int u \;du – \left \{ (\frac{d}{du} \; log \;u) \int u \;du \right \} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \left [\frac{u^{2}}{2} – log\; u – \int \frac{1}{u}. \frac{u^{2}}{2} \;du \right ] \\ = \frac{u^{2}}{2} (log \;u) ^{2} – \frac{u^{2}}{2} (log \;u) + \frac{1}{2} \int u \;du \\ = \frac{u^{2}}{2} (log \;u)^{2} – \frac{u^{2}}{2} (log \;u) + \frac{u^{2}}{4} + C$$

Question 15:

Obtain an integral of (u 2 + 1) log u

Suppose, I = $$\int (u ^{2} + 1) log\; u\; du = \int u ^{2}\; log\; u\; du + \int log\; u\; du \\ Suppose,\; I = I_{1} + I_{2} + …… (1) \\ Where,\; I_{1} = \int u ^{2}\; log\; u\; du \;and\; I_{2} = \int log\; u\; du \\ I_{1} = \int u ^{2}\; log\; u\; du$$

Integrating the equation by parts by taking u as first function and u 2 as second function, we get,

$$I_{1} = (log \;u) – \int u^{2} \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int u ^{2} \;du \right \} \;du \\ = log \;u. \frac{u ^{3}}{3} – \int \frac{1}{u}. \frac{u ^{3}}{3} \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{1}{3} (\int u^{2} \;du) \;du \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} …. (2) \\ I_{2} = \int log\; u\; du$$

Integrating the equation by parts by taking u as first function and u 2 as second function, we get,

$$I_{2} = (log \;u) – \int 1 \;du – \int \left \{ \left ( \frac{d}{du} log \;u \right ) \int 1 \;du \right \} \\ = log \;u. u – \int \frac{1}{u}. u \;du \\ = u\; log \;u – \int 1 \;du \\ = u\; log \;u – u + C_{2} ….. (3)$$

Substituting equations (2) and (3) in equation (1), we get,

$$I = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + C_{1} + u\; log \;u – u + C_{2} \\ = \frac{u ^{3}}{3}\; log \;u – \frac{u ^{3}}{9} + u\; log \;u – u + (C_{1} + C_{2}) \\ = \left (\frac{u ^{3}}{3} + u \right ) log \;u – \frac{u ^{3}}{9} – C$$

Question 16:

Obtain an integral of e u (sin u + cos u)

Answer 16: Based on formulae given in Integrals

Suppose, I = $$\int e^{u} (sin\; u + cos\; u) \; du \\ Suppose,\; f (u) = sin\; u \\ f’ (u) = cos\; u \\ I = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e^{u} f (u) + C \\ I = e^{u} \;sin\; u + C$$

Question 17:

Obtain an integral of $$\frac{e^{u}}{(1 + u) ^{2}}$$

Answer 17: Based on formulae given in Integrals

Suppose, I = $$\int \frac{u \;e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ \frac{u}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1 + u – 1}{(1 + u) ^{2}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{1 + u} – \frac{1}{(1 + u) ^{2}} \right \} \;du \\ Suppose,\; f (u) = \frac{1}{1 + u}, \; \; \; f’ (u) = \frac{- 1}{(1 + u) ^{2}} \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du \\ As\; we\; know,\; \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u} \;f (u) + C \\ \int \frac{u\; e^{u}}{(1 + u) ^{2}} \;du = \frac{e^{u}}{1 + u} + C$$

Question 18:

Obtain an integral of $$e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right )$$

Answer 18: Based on formulae given in Integrals

$$e^{u} \left ( \frac{1 + sin\; u}{1 + cos\; u} \right ) \\ = e^{u} \left ( \frac{sin ^{2} \;\frac{u}{2} + cos ^{2} \;\frac{u}{2} + 2 sin \;\frac{u}{2} \;cos\; \frac{u}{2}}{2\; cos ^{2} \;\frac{u}{2}} \right ) \\ = \frac{e^{u} \left ( sin \;\frac{u}{2} + cos \;\frac{u}{2} \right ) ^{2}}{2 cos ^{2} \;\frac{u}{2}} \\ = \frac{1}{2} e^{u} \left ( \frac{sin \;\frac{u}{2} + cos \;\frac{u}{2}}{cos \;\frac{u}{2}} \right ) ^{2} \\ = \frac{1}{2} e^{u} \left [ tan \;\frac{u}{2} + 1 \right ] ^{2} \\$$ $$= \frac{1}{2} e^{u} \left [1 + tan \;\frac{u}{2} \right ] ^{2} \\ = \frac{1}{2} e^{u} \left [ 1 + tan ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ = \frac{1}{2} e^{u} \left [ sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \\ \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = \left [ \frac{1}{2} sec ^{2} \;\frac{u}{2} + 2 tan\; \frac{u}{2} \right ] \; \; \; ….. (1) \\ Suppose,\; tan\; \frac{u}{2} = f (u) \; so \; f’ (u) = \frac{1}{2} sec ^{2} \;\frac{u}{2}$$

As we know,

$$\int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C$$

Considering equation (1), we get,

$$\int \frac{e^{u} (1 + sin\; u) \;du}{(1 + cos\; u)} = e^{u} tan\; \frac{u}{2} + C$$

Question 19:

Obtain an integral of $$e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right )$$

Answer 19: Based on formulae given in Integrals

Assume, I = $$\int e ^{u} \left ( \frac{1}{u} – \frac{1}{u ^{2}} \right ) \; du \\ Suppose,\; \frac{1}{u} = f (u) \; \; \; f’ (u) = \frac{- 1}{u^{2}} \\ As\; we\; know,\\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ I = \frac{e ^{u}}{u} + C$$

Question 20:

Obtain an integral of $$\frac{(u – 3) e^{u}}{(u – 1) ^{3}}$$

Answer 20: Based on formulae given in Integrals

$$\int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \int e^{u} \left \{ \frac{(u – 3)}{(u – 1 – 2) ^{3}} \right \} \;du \\ = \int e^{u} \left \{ \frac{1}{(u – 1) ^{2} – \frac{2}{(u – 1) ^{3}}} \right \} \;du \\ f (u) = \frac{1}{(u – 1) ^{2}} \; \; \; f’ (u) = \frac{- 2}{(u – 1) ^{3}} \\ As\; we\; know, \\ \int e^{u} \left \{ f (u) + f’ (u) \right \} \;du = e ^{u}\; f (u) + C \\ \int e^{u} \left \{ \frac{(u – 3)}{(u – 1) ^{3}} \right \} \;du = \frac{e ^{u}}{(u – 1) ^{2}} + C$$