The availability of ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals Ex 8.5 encourages students to tackle difficult exercises.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.5

Very Short answer type questions (1 to 3):

Evaluate the following (1 to 23) integrals:

Question 1.
(i) ∫ 2x sin (x2 + 1) dx
(ii) ∫ x3 cos (x4) dt
Solution:
(i) Let I = ∫ 2x sin (x2 + 1) dx
put x2 + 1 = t
⇒ 2x dx = dt
∴ I = ∫ sin t dt
= – cos t + C
= – cos (x2 + 1) + C

(ii) Let I = ∫ x3 cos (x4) dt
put x4 = t
⇒ d (x4) = dt
⇒ 4x3 dx = dt
⇒ x3 dx = $$\frac{1}{4}$$ dt
∴ I = ∫ cos x4 (x3 dx)
= ∫ cos t . $$\frac{d t}{4}$$
= $$\frac{1}{4}$$ sin t + c
= $$\frac{1}{4}$$ sin x4 + c

Question 2.
(i) ∫ $$\frac{\cos \sqrt{x}}{\sqrt{x}}$$ dx
(ii) ∫ $$\frac{\sin \sqrt{x}}{\sqrt{x}}$$ dx
(iii) ∫ $$\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}$$ dx
(iv) ∫ $$\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\cos \sqrt{x}}{\sqrt{x}}$$ dx
put √x = t
⇒ $$\frac{1}{2 \sqrt{x}}$$ dx = dt
⇒ $$\frac{d x}{\sqrt{x}}$$ = 2 dt
∴ I = ∫ cos t (2 dt) = 2 sin t + C

(ii) Let I = ∫ $$\frac{\sin \sqrt{x}}{\sqrt{x}}$$ dx
put √x = t
⇒ $$\frac{1}{\sqrt{x}}$$ dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos √x + c

(iii) Let I = ∫ $$\frac{\sec ^2 \sqrt{x}}{\sqrt{x}}$$ dx
put √x = t
⇒ d (√x) = dt
⇒ $$\frac{1}{2 \sqrt{x}}$$ dx = dt
⇒ $$\frac{d x}{\sqrt{x}}$$ = 2 dt
∴ I = 2 ∫ sec2 t dt
= 2 tan t + c
= 2 tan √x + c

(iv) Let I = ∫ $$\frac{\ {cosec}^2 \sqrt{x}}{\sqrt{x}}$$ dx
put √x = t
⇒ $$\frac{1}{2 \sqrt{x}}$$ dx = dt
∴ I = ∫ cosec2 t (2 dt) = – 2 cot t + C
= – 2 cot √x + C.

Question 3.
(i) ∫ sin x . ecos x dx
(ii) ∫ $$\frac{e^{2 x}}{1+e^x}$$ dx
Solution:
(i) Let I = ∫ sin x . ecos x dx
put cos x = t
⇒ – sin x dx = dt
= ∫ et (- dt)
= – et + C
= – ecos x + C

(ii) Let I = ∫ $$\frac{e^{2 x}}{1+e^x}$$ dx
= ∫ $$\frac{e^x \cdot e^x d x}{1+e^x}$$ dx
put ex = t
⇒ d (ex) = dt
⇒ ex dx = dt
= ∫ $$\frac{t d t}{1+t}$$
= ∫ $$\frac{1+t-1}{1+t}$$
= ∫ $$\left[1-\frac{1}{t+1}\right]$$ dt
= t – log |1 + t| + c
= ex – log |1 + ex| + c

Question 4.
(i) ∫ sin x sin (cos x) dx (NCERT)
(ii) ∫ x3 sec2 (x4 + 3) dx
Solution:
(i) Let I = ∫ sin x sin (cos x) dx
put cos x = t
⇒ – sin x dx = dt
∴ I = ∫ sin t (- dt)
= cos t + C
= cos (cos x) + C

(ii) Let I = ∫ x3 sec2 (x4 + 3) dx
put x4 + 3 = t
⇒ 4x3 dx = dt
= ∫ sec2 t ($$\frac{d t}{4}$$) = $$\frac{1}{4}$$ tan t + C
= $$\frac{1}{4}$$ tan (x4 + 4) + C

Question 5.
(i) ∫ $$\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)$$ dx
(ii) ∫ $$\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)$$ dx (ISC 2017)
(iii) ∫ 2x sec3 (x2 + 5) tan (x2 + 5) dx
Solution:
(i) Let I = ∫ $$\frac{1}{x^2} \tan ^2\left(\frac{1}{x}\right)$$ dx
put $$\frac{1}{x}$$ = t
⇒ – $$\frac{1}{x^2}$$ dx = dt
∴ I = ∫ tan2 t (- dt)
= ∫ – (sec2 t dt + ∫ dt + C
= – tan t + t + C
= – tan ($$\frac{1}{x}$$) + $$\frac{1}{x}$$ + C

(ii) Let I = ∫ $$\frac{1}{x^2} \sin ^2\left(\frac{1}{x}\right)$$ dx

(iii) Let I = ∫ 2x sec3 (x2 + 5) tan (x2 + 5) dx
put x2 + 5 = t
⇒ 2x dx = dt
= ∫ sec3 t tan t dt
= ∫ sec2 t (sec t tan t dt)
= $$\frac{\sec ^3 t}{3}$$ + C
[∵ [f(x)]n f'(x) dx = $$\frac{[f(x)]^{n+1}}{n+1}$$ + C, where n ≠ – 1]
= $$\frac{1}{3}$$ sec3 (x2 + 5) + C

Question 6.
(i) ∫ $$\frac{\sin (\log x)}{x}$$ dx
(ii) ∫ $$\frac{\ {cosec}^2(\log x)}{x}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\sin (\log x)}{x}$$ dx
put log x = t
⇒ d (log x) dt
⇒ $$\frac{1}{x}$$ dx = dt
∴ I = ∫ sin t dt
= – cos t + c
= – cos (log x) + c

(ii) Let I = ∫ $$\frac{\ {cosec}^2(\log x)}{x}$$ dx
put log x = t
⇒ $$\frac{1}{x}$$ dx = dt
∴ I = ∫ cosec2 t dt
= – cot t + C
= – cot (log x) + C

Question 7.
(i) ∫ $$\frac{e^{\tan ^{-1} x}}{1+x^2}$$ dx
(ii) ∫ $$\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{e^{\tan ^{-1} x}}{1+x^2}$$ dx
put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
= ∫ et dt
= et + C
= etan-1 x + C

(ii) Let I = ∫ $$\frac{\sin \left(2 \tan ^{-1} x\right)}{1+x^2}$$ dx
put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
∴ I = ∫ sin 2t dt
= – $$\frac{\cos 2 t}{2}$$ + C
= – $$\frac{1}{2}$$ cos (2 tan-1 x) + C

Question 8.
(i) ∫ $$\frac{e^{-1 / x}+1}{x^2}$$ dx
(ii) ∫ $$\frac{e^{m \tan ^{-1} x}}{1+x^2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{e^{-1 / x}+1}{x^2}$$ dx
= ∫ (e-1/x + 1) $$\frac{1}{x^{2}}$$ dx
put – $$\frac{1}{x}$$ = t
⇒ $$\frac{1}{x^{2}}$$ dx = dt
= ∫ (et + 1) dt
= et + t + C
= e– 1/x – $$\frac{1}{x}$$ + C

(ii) Let I = ∫ $$\frac{e^{m \tan ^{-1} x}}{1+x^2}$$ dx
put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
∴ I = ∫ emt dt
= $$\frac{e^{m t}}{m}$$ + C
= $$\frac{e^{m \tan ^{-1} x}}{m}$$ + C

Question 9.
(i) ∫ ex cosec2 (ex) dt
(ii) ∫ $$\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}$$ dx
Solution:
(i) Let I = ∫ ex cosec2 (ex) dt
put ex = t
⇒ ex dx = dt
∴ I = ∫ cosec2 t dt
= – cot t + C
= – cot (ex) + C

(ii) Let I = ∫ $$\frac{(x+1) e^x}{\cot ^2\left(x e^x\right)}$$ dx
put x ex = t
d (x ex) = dt
⇒ (x ex + ex) dx = dt
⇒ (x + 1) ex dx = dt
∴ I = ∫ $$\int \frac{d t}{\cot ^2 t}$$
= ∫ tan2 t dt
= ∫ (sec2 t – 1) dt
= tan t – t + C
= tan (x ex) – x ex + C

Question 10.
(i) ∫ x2 ex3 (sin ex3) dx
(ii) ∫ $$\frac{\sin ^2(\log x)}{x}$$ dx
(iii) ∫ $$\frac{\sqrt{\tan x}}{\sin ^2 x}$$ dx
Solution:
(i) Let I = ∫ x2 ex3 (sin ex3) dx
put ex3 = t
⇒ ex3 . 3x2 dx = dt
∴ I = $$\frac{1}{3}$$ ∫ sin t dt
= – $$\frac{\cos t}{3}$$ + C
= – $$\frac{1}{3}$$ cos (ex3) + C

(ii) Let I = ∫ $$\frac{\sin ^2(\log x)}{x}$$ dx
put log x = t
⇒ $$\frac{1}{x}$$ dx = dt
∴ I = ∫ sin2 t dt
= ∫ $$\frac{1-\cos 2 t}{2}$$ dt
= $$\frac{1}{2}\left[t-\frac{\sin 2 t}{2}\right]$$ + C
= $$\frac{1}{2}$$ [log x – $$\frac{1}{2}$$ sin (2 log x)] + C

(iii) Let I = ∫ $$\frac{\sqrt{\tan x}}{\sin ^2 x}$$ dx
= ∫ $$\frac{\sqrt{\tan x}}{\frac{\sin ^2 x}{\cos ^2 x} \cos ^2 x}$$ dx
= ∫ $$\frac{\sqrt{\tan x}}{\tan ^2 x}$$ sec2 x dx
= ∫ (tan x)-3/2 sec2 x dx
put tan x = t
⇒ sec2 x dx = dt
= ∫ t-3/2 dt
= $$\frac{t^{-3 / 2}+1}{-\frac{3}{2}+1}$$ + C
= – 2 t-1/2 + C
= – $$\frac{2}{\sqrt{\tan x}}$$ + C

Question 11.
(i) ∫ x sin3 (x2) cos (x2) dx
(ii) ∫ tan 2x sec 2x dx (NCERT)
Solution:
(i) Let I = ∫ x sin3 (x2) cos (x2) dx
put sin x2 = t
⇒ cos (x2) . 2x dx = dt
= ∫ t3 $$\frac{d t}{2}$$
= $$\frac{1}{2}$$ $$\frac{t^4}{4}$$ + C
= $$\frac{t^4}{8}$$ + C
= $$\frac{1}{8}$$ sin4 (x2 + C

(ii) Let I = ∫ tan3 2x sec 2x dx
= ∫ tan2 2x (tan 2x sec 2x) dx
= ∫ (sec2 2x – 1) tan 2x sec 2x dx
put sec 2x = t
⇒ (sec 2x tan 2x) 2 dx = dt
= ∫ (t2 – 1) $$\frac{d t}{2}$$
= $$\frac{1}{2}\left[\frac{t^3}{3}-t\right]$$ + C
= $$\frac{t^3}{6}-\frac{1}{2} t$$ + C
= $$\frac{1}{6}$$ sec3 2x – $$\frac{1}{2}$$ sec 2x + C

Question 12.
(i) ∫ sin3 (2x + 1) dx (NCERT)
(ii) ∫ sin3 x cos2 x dx (NCERT)
Solution:
(i) Let I = ∫ sin3 (2x + 1) dx
= ∫ sin2 (2x + 1) sin (2x + 1) dx
= ∫ (1 – cos2 (2x + 1)) sin (2x + 1) dx
put cos (2x + 1) = t
⇒ – 2 sin (2x + 1) dx = dt
∴ I = ∫ (1 – t2) $$\frac{d t}{- 2}$$
=  + C
= – $$\frac{1}{2}$$ cos (2x + 1) + $$\frac{1}{6}$$ cos3(2x + 1) + C

(ii) Let I = ∫ sin3 x cos2 x dx
= ∫ $$\frac{\cos ^3 x}{\sin ^2 x}$$ dx
= ∫ $$\frac{\left(1-\sin ^2 x\right) \cos x}{\sin ^2 x}$$ dx
put sin x = t
⇒ cos x dx = dt
∴ I = ∫ $$\frac{\left(1-t^2\right) d t}{t^2}$$
= ∫ $$\frac{1}{t^2}$$ dt – ∫ dt
= – $$\frac{1}{t}$$ – t + C
= – $$\frac{1}{sin x}$$ – sin x + C
= – cosec x – sin x + C

Question 14.
(i) ∫ $$\frac{\cos ^3 x}{\sqrt{\sin x}}$$ dx
(ii) ∫ $$\frac{\sin x \cos ^3 x}{1+\cos ^2 x}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\cos ^3 x}{\sqrt{\sin x}}$$ dx
= ∫ $$\frac{\cos ^2 x \cos x d x}{\sqrt{\sin x}}$$
= ∫ $$\frac{\left(1-\sin ^2 x\right) \cos x d x}{\sqrt{\sin x}}$$
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{1-t^2}{\sqrt{t}}$$ dt
= ∫ $$\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]$$ dt
put sin x = t
⇒ cos x dx = dt
= ∫ $$\frac{1-t^2}{\sqrt{t}}$$ dt
= ∫ $$\left[\frac{1}{\sqrt{t}}-t^{3 / 2}\right]$$ dt
= $$\frac{t^{\frac{-1}{2}+1}}{\left(\frac{-1}{2}+1\right)}-\frac{t^{\frac{3}{2}+1}}{\left(\frac{3}{2}+1\right)}$$ + c
= 2√t – $$\frac{2}{5}$$ t5/2 + c
= 2 $$\sqrt{sin x}$$ – $$\frac{2}{5}$$ t5/2 + c

(ii) Let I = ∫ $$\frac{\sin x \cos ^3 x}{1+\cos ^2 x}$$ dx
put cos x = t
⇒ – sin x dx = dt

Question 15.
(i) ∫ cosec x log (cosec x – cot x) dx
(ii) ∫ $$\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}$$ dx
Solution:
(i) Let I = ∫ cosec x log (cosec x – cot x) dx
put log (cosec x – cot x) = t
⇒ d {log (cosec x – cot x)} = dt
[- cot x cosec x + cosec2 x] dx = dt
⇒ cosec x dx = $$\frac{1}{\ {cosec} x-\cot x}$$ dt
∴ I = ∫ t dt
= $$\frac{t^{2}}{2}$$ + c
= $$\frac{1}{2}$$ [log (cosec x – cot x)]2 + c

(ii) Let I = ∫ $$\frac{\log \left(\tan \frac{x}{2}\right)}{\sin x}$$ dx
put log (tan $$\frac{x}{2}$$) = t
⇒ $$\frac{1}{\tan \frac{x}{2}} \sec ^2 \frac{x}{2} \cdot \frac{1}{2}$$ dx = dt
⇒ $$\frac{1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$$ dx = dt
⇒ $$\frac{1}{\sin x}$$ dx = dt
∴ I = ∫ t dt
= $$\frac{t^{2}}{2}$$ + c
= $$\frac{\left[\log \left(\tan \frac{x}{2}\right)\right]^2}{2}$$ + C

Question 16.
(i) ∫ sec4 x dx
(ii) ∫ tan2 x sec4 x dx (NCERT Exemplar)
Solution:
(i) Let I = ∫ sec4 x dx
= ∫ sec2 x (1 + tan2 x) dx
put tan x = t
⇒ sec2 x dx = dt
∴ I = ∫ (1 + t2 dt
= t + $$\frac{t^{3}}{3}$$ + C
= tan x + $$\frac{\tan ^3 x}{3}$$ + C

(ii) Let I = ∫ tan2 x sec4 x dx
= ∫ tan2 x . sec2 x . sec2 x dx
= ∫ tan2 x (1 + tan2 x) . sec2 x dx
put tan x = t
sec2 x dx = dt
= ∫ t2 (1 + t2) dt
= $$\frac{t^3}{3}+\frac{t^5}{5}$$ + C
= $$\frac{1}{3}$$ tan3 x + $$\frac{1}{5}$$ tan5 x + C

Question 17.
(i) ∫ $$\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}$$ dx
(ii) ∫ $$\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\tan ^5 \sqrt{x} \sec ^2 \sqrt{x}}{\sqrt{x}}$$ dx
put tan √x = t
sec2 √x $$\left(\frac{1}{2 \sqrt{x}}\right)$$ dx = dt
∴ I = ∫ t5 (2 dt)
= $$\frac{2 t^6}{6}$$ + C
= $$\frac{1}{3}$$ tan6 (√x) + C

(ii) Let I = ∫ $$\frac{\cos x-\sin x}{\sqrt{1+\sin 2 x}}$$ dx
= ∫ $$\frac{(\cos x-\sin x) d x}{\sqrt{\cos ^2 x+\sin ^2 x+2 \sin x \cos x}}$$
= ∫ $$\frac{(\cos x-\sin x) d x}{\sqrt{(\cos x+\sin x)^2}}$$
put cos x + sin x = t
⇒ (- sin x + cos x) dx = dt
= ∫ $$\frac{d t}{t}$$
= log |t| + C
= log |cos x + sin x| + C

Question 18.
(i) ∫ $$\frac{x}{\sqrt{x+4}}$$ dx (NCERT)
(ii) ∫ x $$\sqrt{x+2}$$ dx (NCERT)
Solution:
(i) Let I = ∫ $$\frac{x d x}{\sqrt{x+4}}$$
= ∫ $$\frac{(x+4-4)}{\sqrt{x+4}}$$ dx
= ∫ $$\sqrt{x+4}$$ dx – 4 ∫ (x + 4)$$-\frac{1}{2}$$ dx
= $$\frac{2}{3}$$ (x + 4)$$\frac{3}{2}$$ – 4 $$\frac{(x+4)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)}$$ + C
= $$\frac{2}{3}$$ (x + 4)$$\frac{3}{2}$$ – 8 $$\sqrt{x+4}$$ + C
∴ I = $$\frac{2}{3} \sqrt{x+4}$$ (x + 4 – 12)
= $$\frac{2}{3}$$ (x – 8) $$\sqrt{x+4}$$ + C

(ii) Let I = ∫ x $$\sqrt{x+2}$$ dx
= ∫ (x + 2 – 2) $$\sqrt{x+2}$$ dx
= ∫ (x + 2)$$\frac{3}{2}$$ dx – 2 ∫ (x + 2)$$\frac{1}{2}$$ dx
= $$\frac{2}{5}(x+2)^{\frac{5}{2}}-2 \times \frac{2}{3}(x+2)^{\frac{3}{2}}$$ + C
= $$\frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}$$ + C

Question 19.
(i) ∫ $$\frac{x}{\left(x^2+1\right)^2}$$ dx
(ii) ∫ $$\frac{x^2}{(4+x)^{3 / 2}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x}{\left(x^2+1\right)^2}$$ dx
put x2 = t
⇒ 2x dx = dt
∴ I = ∫ $$\frac{d t}{2(t+1)^2}$$
= $$\frac{1}{2} \frac{(t+1)^{-2+1}}{(-2+1)}$$ + C
= – $$\frac{1}{2(t+1)}$$ + C
= – $$\frac{1}{2\left(x^2+1\right)}$$ + C

(ii) Let I = ∫ $$\frac{x^2}{(4+x)^{3 / 2}}$$ dx
put 4 + x = t
⇒ dx = dt

Question 20.
(i) ∫ 4x3 $$\sqrt{5-x^2}$$ dx
(ii) ∫ $$\frac{d x}{x-\sqrt{x}}$$ (NCERT)
Solution:
(i) Let I = ∫ 4x3 $$\sqrt{5-x^2}$$ dx
put 5 – x2 = t
⇒ x2 = 5 – t
⇒ 2x dx = – dt
∴ I = ∫ 2 (5 – t) √t (- dt)
= – 10 t$$\frac{1}{2}$$ + 1 / ($$\frac{1}{2}$$ + 1) + 2 t$$\frac{5}{2}$$ / $$\frac{5}{2}$$ + C
= – $$\frac{20}{3}$$ t3/2 + $$\frac{4}{5}$$ t5/2 + C
= $$\frac{4}{5}$$ (5 – x2)5/2 – $$\frac{20}{3}$$ (5 – x2)5/2 + C

(ii) Let I = ∫ $$\frac{d x}{x-\sqrt{x}}$$
put √x = t
⇒ x = t2
⇒ dx = 2t dt
∴ I = ∫ $$\frac{2 t d t}{t^2-t}$$
= ∫ $$\frac{2 t d t}{t(t-1)}$$
= 2 ∫ $$\frac{d t}{t-1}$$
= 2 log |t – 1| + C
= 2 log |√x – 1| + C

Question 21.
(i) ∫ $$\frac{x}{1+\sqrt{x}}$$ dx (NCERT Exemplar)
(ii) ∫ $$\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x}{1+\sqrt{x}}$$ dx
put √x = t
⇒ x = t2
⇒ dx = 2t dt

(ii) Let I = ∫ $$\frac{x^{\frac{1}{2}}}{1+x^{\frac{3}{4}}}$$ dx
put x1/4 = t
⇒ x = t4
⇒ dx = 4t3

Question 22.
(i) ∫ $$\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}$$
(ii) ∫ $$\frac{d x}{\sqrt{1+\sqrt{x}}}$$
Solution:
(i) Let I = ∫ $$\frac{d x}{\sqrt{x+1}+\sqrt[3]{x+1}}$$

(ii) Let I = ∫ $$\frac{d x}{\sqrt{1+\sqrt{x}}}$$
put $$\sqrt{1+\sqrt{x}}$$ = t
⇒ 1 + √x = t
⇒ √x = t2 – 1
⇒ x = (t2 – 1)2
⇒ dx = 4t (t2 – 1)
∴ I = ∫ $$\frac{4 t\left(t^2-1\right)}{t}$$
= 4 $$\left[\frac{t^3}{3}-t\right]$$ + C
= $$\frac{4}{3}$$ (1 + √x)3/2 – 4 $$\sqrt{1+\sqrt{x}}$$ + C

Question 23.
(i) ∫ $$\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}$$ dx
(ii) ∫ $$\frac{\sqrt{1+x^2}}{x^4}$$ dx (NCERT Exemplar)
(iii) ∫ $$\frac{1}{x^2 \sqrt{1+x^2}}$$ dx
Solution:
(i) Let I = ∫ $$\frac{\sin 2 x}{\left(a^2+b^2 \sin ^2 x\right)^2}$$ dx
put a2 + b2 sin2 x = t
⇒ 2b2 sin x cos x dx = dt
⇒ b2 sin 2x dx = dt

(ii) Let I = ∫ $$\frac{\sqrt{1+x^2}}{x^4}$$ dx
put x = tan θ
dx = sec2 θ
∴ I = ∫ $$\frac{\sqrt{1+\tan ^2 \theta}}{\tan ^4 \theta}$$ sec2 θ dθ
= ∫ $$\frac{\sec ^3 \theta d \theta}{\tan ^4 \theta}$$
= ∫ $$\frac{\frac{1}{\cos ^3 \theta}}{\frac{\sin ^4 \theta}{\cos ^4 \theta}}$$ dθ
= ∫ $$\frac{\cos \theta d \theta}{\sin ^4 \theta}$$
= ∫ (sin θ)-4 cos θ dθ
= $$\frac{(\sin \theta)^{-4+1}}{(-4+1)}$$ + C
[∵ ∫ [f(x)]n f'(x) dx = $$\frac{[f(x)]^{n+1}}{n+1}$$ + C]

(iii) Let I = ∫ $$\frac{1}{x^2 \sqrt{1+x^2}}$$ dx

Question 24.
(i) If ∫ x ekx2 dx = $$\frac{1}{4}$$ e2x2 + C, then find the value of k.
(ii) If ∫ x6 sin (5x)7 dx = $$\frac{k}{5}$$ cos (5x7) + C, then what is the value of k ?
Solution:
(i) Let I = ∫ x ekx2 dx
put x2 = t
2x dx = dt
= ∫ ekt $$\frac{d t}{2}$$
= $$\frac{1}{2} \frac{e^{k t}}{k}$$ + C
= $$\frac{1}{2 k}$$ ekx2 + C ……….(1)
Also given
I = ∫ x ekx2 dx
= $$\frac{1}{4}$$ e2x2 + C …………..(2)
From (1) and (2) ; we have

⇒ 2k = 4
⇒ k = 2

(ii) put x7 = t
⇒ 7x6 dx = dt
⇒ x6 dx = $$\frac{d t}{7}$$
∴ ∫ x6 sin (5x7) dx = ∫ sin 5t $$\frac{d t}{7}$$
= – $$\frac{\cos 5 x^7}{35}$$ + C ……….(1)
Also given,
∫ x6 sin (5x7) dx = $$\frac{k}{5}$$ cos (5x7) + C ………..(2)
∴ from (1) and (2) ; we have
$$\frac{k}{5}=-\frac{1}{35}$$
k = – $$\frac{1}{7}$$