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

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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 . e^{cos x} dx
(ii) ∫ \(\frac{e^{2 x}}{1+e^x}\) dx
Solution:
(i) Let I = ∫ sin x . e^{cos x} dx
put cos x = t
⇒ – sin x dx = dt
= ∫ e^t (- dt)
= – e^t + C
= – e^{cos 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 e^x = t
⇒ d (e^x) = dt
⇒ e^x 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
= e^x – log |1 + e^x| + 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
= ∫ e^t dt
= e^t + C
= e^{tan-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
= ∫ (e^t + 1) dt
= e^t + 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 = ∫ e^mt dt
= \(\frac{e^{m t}}{m}\) + C
= \(\frac{e^{m \tan ^{-1} x}}{m}\) + C

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

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

Question 10.
(i) ∫ x² e^x3 (sin e^x3) dx
(ii) ∫ \(\frac{\sin ^2(\log x)}{x}\) dx
(iii) ∫ \(\frac{\sqrt{\tan x}}{\sin ^2 x}\) dx
Solution:
(i) Let I = ∫ x² e^x3 (sin e^x3) dx
put e^x3 = t
⇒ e^x3 . 3x² dx = dt
∴ I = \(\frac{1}{3}\) ∫ sin t dt
= – \(\frac{\cos t}{3}\) + C
= – \(\frac{1}{3}\) cos (e^x3) + 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 – \(\frac{2}{5}\) t^5/2 + c
= 2 \(\sqrt{sin x}\) – \(\frac{2}{5}\) t^5/2 + 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
= – \(\frac{20}{3}\) t^3/2 + \(\frac{4}{5}\) t^5/2 + 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 x^1/4 = 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 e^kx2 dx = \(\frac{1}{4}\) e^2x2 + 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 e^kx2 dx
put x² = t
2x dx = dt
= ∫ e^kt \(\frac{d t}{2}\)
= \(\frac{1}{2} \frac{e^{k t}}{k}\) + C
= \(\frac{1}{2 k}\) e^kx2 + C ……….(1)
Also given
I = ∫ x e^kx2 dx
= \(\frac{1}{4}\) e^2x2 + 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 = – \(\frac{1}{7}\)