Peer review of Class 12 ISC Maths Solutions Chapter 8 Integrals Ex 8.3 can encourage collaborative learning.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.3

Very short answer type questions (1 to 10):

Find all the anti-derivatives of the following (1 to 6) functions :

Question 1.
(i) (ax + b)3
(ii) 4e3x + 1 (NCERT)
Solution:
(i) ∫ (ax + b)3 = $$\frac{(a x+b)^{3+1}}{(3+1) a}$$ + C
= $$\frac{(a x+b)^4}{4 a}$$ + C
[∵ ∫ (ax + b)n dx = $$\frac{(a x+b)^{n+1}}{(n+1) a}$$ + C ; n ≠ – 1]

(ii) ∫ 4e3x + 1 dx
= 4 ∫ e3x dx + ∫ dx
= $$\frac{4 e^{3 x}}{3}$$ + x + C

Question 1 (old).
(i) sin 2x (NCERT)
(ii) cos 3x (NCERT)
Solution:
(i) ∫ sin 2x dx = – $$\frac{\cos 2 x}{2}$$ + C
[∵ ∫ sin mx dx = $$\frac{\cos m x}{m}$$ + C]

(ii) ∫ cos 3x dx = $$\frac{\sin 3 x}{3}$$ + C

Question 2.
(i) $$\frac{1}{\sqrt{7-2 x}}$$
(ii) $$\sqrt{a x+b}$$ (NCERT)
Solution:
(i) ∫ $$\frac{1}{\sqrt{7-2 x}}$$ dx
= ∫ (7 – 2x)– $$\frac{1}{2}$$ dx
= $$\frac{(7-2 x)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)(-2)}$$ + C
= – $$\sqrt{7-2x}$$ + C

(ii) ∫ $$\sqrt{a x+b}$$ dx
= ∫ (ax + b)1/2 dx
= $$\frac{(a x+b)^{\frac{1}{2}+1}}{\left(\frac{1}{2}+1\right) a}$$
= $$\frac{2}{3 a}$$ (ax + b)3/2 + C

Question 2 (old).
(i) sin mx (NCERT)
(ii) e2x (NCERT)
Solution:
(i) ∫ sin mx dx = – $$\frac{\cos m x}{m}$$ + C

(ii) ∫ e2x dx
= $$\frac{e^{2 x}}{2}$$ + C
[∵ ∫ emx dx = $$\frac{e^{m x}}{m}$$ + C]

Question 3.
(i) tan2 (2x – 3) (NCERT)
(ii) 5 cosec2 (2 – 7x)
Solution:
(i) ∫ tan2 (2x – 3) dx
= ∫ [sec2 (2x – 3) – 1] dx
= ∫ [sec2 (2x – 3) dx – ∫ dx
= $$\frac{\tan (2 x-3)}{2}$$ – x + c

(ii) ∫ 5 cosec2 (2 – 7x) dx
= $$\frac{-5 \cot (2-7 x)}{-7}$$ + C
= $$\frac{5}{7}$$ cot (2 – 7x) + C

Question 4.
(i) $$\sqrt{e^x}$$
(ii) cosec (3x + 2) cot (3x + 2)
Solution:
(i) ∫ $$\sqrt{e^x}$$ dx
= ∫ ex/2 dx
= $$\frac{e^{x / 2}}{1 / 2}$$ + C
= 2ex/2 + C

(ii) ∫ cos (3x + 2) cot (3x + 2) dx
= – $$\frac{\ {cosec}(3 x+2)}{3}$$ + C
[∵ ∫ cot θ cosec θ dθ = – cosec θ + C]

Evaluate the following (7 to 21) integrals :

Question 5.
(i) ∫ e2x + 3 dx
(ii) ∫ e(3a log x + e3x log a) dx
Solution:
(i) ∫ e2x + 3 dx
= $$\frac{e^{2 x+3}}{2}$$ + C
[∵ ∫ emx dx = $$\frac{e^{m x}}{m}$$ + C]

(ii) ∫ e(3a log x + e3x log a) dx
= ∫ (elog x3a + elog a3x) dx
[∵ a log b = log ba
and elog x = x]
= ∫ (x3a + a) dx
= $$\frac{x^{3 a+1}}{3 a+1}+\frac{a^{3 x}}{3 \log a}$$ + C

Question 6.
(i) ∫ $$\sqrt{1+cos x}$$ dx
(ii) ∫ $$\sqrt{1+sin x}$$ dx
Solution:
(i) ∫ $$\sqrt{1+cos x}$$ dx
= ∫ $$\sqrt{2 \cos ^2 \frac{x}{2}}$$ dx
= √2 ∫ cos $$\frac{x}{2}$$ dx
= √2 $$\frac{\sin \frac{x}{2}}{\frac{1}{2}}$$ + C
= 2√2 sin $$\frac{x}{2}$$ + C

(ii) ∫ $$\sqrt{1+sin x}$$ dx

Question 6 (old).
(ii) ∫ $$\sqrt{1-sin x}$$ dx
Solution:

Question 7.
(i) ∫ sec2 (7 – x) dx
(ii) ∫ a3x + 5 dx, a ≥ 0
Solution:
(i) ∫ sec2 (7 – x) dx
= ∫ sec2 t (- dt)
[put 7 – x = t
⇒ dx = – dt]
= – tan t + C
= – tan (7 – x) + C

(ii) ∫ a3x + 5 dx, a > 0
= $$\frac{a^{3 x+5}}{3(\log a)}$$ + C
[∵ ∫ amx dx = $$\frac{a^{m x}}{m \log a}$$ + C

Question 8.
(i) ∫ $$\frac{1-\tan ^2 x}{1+\tan ^2 x}$$ dx
(ii) ∫ sec2 x cosec2 x dx
Solution:
(i) ∫ $$\frac{1-\tan ^2 x}{1+\tan ^2 x}$$ dx
= ∫ cos 2x dx
= $$\frac{\sin 2 x}{2}$$ + C

(ii) ∫ sec2 x cosec2 x dx
= ∫ $$\frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin ^2 x \cos ^2 x}$$
= ∫ $$\frac{1}{\cos ^2 x}$$ dx + ∫ $$\frac{1}{\sin ^2 x}$$ dx
= ∫ sec2 x dx + ∫ cosec2 x dx
= tan x – cot x + C

Question 9.
(i) ∫ $$\frac{x}{2 x-3}$$ dx
(ii) ∫ $$\frac{x+2}{3 x^2+5 x-2}$$ dx
Solution:
(i) Let I = ∫ $$\frac{x}{2 x-3}$$ dx
= $$\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]$$ dx
= $$\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]$$ dx
= $$\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]$$ + C
= $$\frac{x}{2}$$ – $$\frac{3}{4}$$ log |2x – 3| + C

(ii) Let I = ∫ $$\frac{x+2}{3 x^2+5 x-2}$$ dx
= ∫ $$\frac{(x+2) d x}{(x+2)(3 x-1)}$$
= ∫ $$\frac{d x}{3 x-1}$$
= $$\frac{\log |3 x-1|}{3}$$ + C

Question 10.
(i) ∫ $$\frac{x}{(x+1)^2}$$ dx (ISC 2010)
(ii) ∫ $$\frac{x}{(x+1)^2}$$ dx (NCERT Exemplar)
Solution:
(i) ∫ $$\frac{x}{(x+1)^2}$$ dx

(ii) ∫ $$\frac{x}{(x+1)^2}$$ dx
= ∫ $$\left[\frac{x^2-1+3}{x+1}\right]$$ dx
= ∫ [(x – 1) + $$\frac{3}{x+1}$$] dx
= $$\frac{x^{2}}{2}$$ – x + 3 log |x + 1| + C

Question 11.
(i) ∫ $$\frac{x^3}{2 x+1}$$ dx
(ii) ∫ $$\frac{x^3+4 x^2-3 x-2}{x+2}$$ dx
Solution:
(i) ∫ $$\frac{x^3}{2 x+1}$$ dx

(ii) ∫ $$\frac{x^3+4 x^2-3 x-2}{x+2}$$ dx
= ∫ [(x2 + 2x – 7) + $$\frac{12}{x+2}$$] dx
= $$\frac{x^3}{3}$$ + x2 – 7x + 12 log |x + 2| + C

Question 11 (old).
(i) ∫ $$\frac{1}{(2 x-3)^{3 / 2}}$$ dx
(ii) ∫ $$\frac{x}{2 x-3}$$ dx
Solution:
(i) ∫ $$\frac{1}{(2 x-3)^{3 / 2}}$$ dx
= ∫ (2x – 3)– 3/2 dx
= $$\frac{(2 x-3)^{-\frac{3}{2}+1}}{\left(-\frac{3}{2}+1\right) \cdot 2}$$
= – $$\frac{1}{\sqrt{2 x-3}}$$ + C

(ii) ∫ $$\frac{x}{2 x-3}$$ dx
= $$\frac{1}{2} \int\left[\frac{2 x-3+3}{2 x-3}\right]$$ dx
= $$\frac{1}{2} \int\left[1+\frac{3}{2 x-3}\right]$$ dx
= $$\frac{1}{2}\left[x-\frac{3 \log |2 x-3|}{2}\right]$$ + C
= $$\frac{x}{2}$$ – $$\frac{3}{4}$$ log |2x – 3| + C

Question 12.
(i) ∫ $$\frac{1}{\sqrt{x+1}+\sqrt{x+2}}$$ dx
(ii) ∫ $$\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}$$ dx
Solution:
(i) ∫ $$\frac{1}{\sqrt{x+1}+\sqrt{x+2}}$$ dx
= ∫ $$\frac{\sqrt{x+1}-\sqrt{x+2}}{(x+1)-(x+2)}$$ dx
= – $$\left[\frac{2}{3}(x+1)^{3 / 2}-\frac{2}{3}(x+2)^{3 / 2}\right]$$ + C

(ii) ∫ $$\frac{1}{\sqrt{2 x+3}-\sqrt{2 x}}$$ dx

Question 13.
(i) ∫ $$\frac{x+1}{\sqrt{2 x-1}}$$ dx
(ii) ∫ x $$\sqrt{3 x-2}$$ dx
Solution:
(i) ∫ $$\frac{x+1}{\sqrt{2 x-1}}$$ dx

(ii) ∫ x $$\sqrt{3 x-2}$$ dx
= $$\frac{1}{3}$$ ∫ (3x – 2 + 2) $$\sqrt{3 x-2}$$ dx
= $$\frac{1}{3}$$ ∫ (3x – 2)3/2 dx + $$\frac{2}{3}$$ ∫ (3x – 2)1/2 dx
= $$\frac{1}{3} \frac{(3 x-2)^{5 / 2}}{\frac{5}{2} \times 3}+\frac{2}{3} \frac{(3 x-2)^{3 / 2}}{\frac{3}{2} \times 3}$$ + C
= $$\frac{2}{45}$$ (3x – 2)5/2 + $$\frac{4}{27}$$ (3x – 2)3/2 + C

Question 14.
(i) ∫ cos2 x dx (NCERT)
(ii) ∫ sin4 dx
Solution:
(i) ∫ cos2 x dx
= ∫ $$\frac{1+\cos 2 x}{2}$$ dx
= $$\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]$$ + C

(ii) ∫ sin4 dx
= ∫ $$\left[\frac{1-\cos 2 x}{2}\right]^2$$ dx
= $$\frac{1}{4}$$ ∫ [1 + cos2 2x – 2 cos 2x] dx
= $$\frac{1}{4}$$ ∫ [1 + $$\frac{1+\cos 4 x}{2}$$ – 2 cos 2x] dx
= $$\frac{1}{8}$$ ∫ [3 + cos 4x – 4 cos 2x] dx
= $$\frac{1}{8}$$ [3x + $$\frac{\sin 4 x}{4}$$ – 2 sin 2x] + C

Question 15.
(i) ∫ sin5 (2x + 5) dx (NCERT)
(ii) ∫ cos4 2x dx (NCERT)
Solution:
(i) ∫ sin5 (2x + 5) dx
= ∫ $$\frac{1-\cos (4 x+10)}{2}$$ dx
[∵ sin2 θ = $$\frac{1-\cos 2 \theta}{2}$$]
= ∫ [x – $$\frac{\sin (4 x+10)}{4}$$] + C

(ii) ∫ cos4 2x dx
= ∫ (cos2 2x)2 dx
= ∫ $$\left[\frac{1+\cos 4 x}{2}\right]^2$$ dx
= $$\frac{1}{4}$$ ∫ [1 + 2 cos 4x + cos2 4x] dx
= $$\frac{1}{4}$$ ∫ [1 + 2 cos 4x + $$\frac{1+\cos 8 x}{2}$$] dx
= $$\frac{1}{8}$$ ∫ [3 + 4 cos 4x + cos 8x] dx
= $$\frac{1}{8}$$ [3x + 4 $$\frac{\sin 4 x}{4}$$ + $$\frac{\sin 8 x}{8}$$] + c
= $$\frac{3 x}{8}+\frac{\sin 4 x}{8}+\frac{\sin 8 x}{64}$$ + c

Question 16.
(i) ∫ $$\frac{1}{1-\sin \frac{x}{2}}$$ dx
(ii) ∫ sin x $$\sqrt{1-cos 2x}$$ dx
Solution:
(i) ∫ $$\frac{1}{1-\sin \frac{x}{2}}$$ dx

(ii) ∫ sin x $$\sqrt{1-cos 2x}$$ dx
= ∫ sin x $$\sqrt{2 \sin ^2 x}$$ dx
= √2 ∫ sin2 x dx
= $$\sqrt{2} \int \frac{1-\cos 2 x}{2}$$ dx
= $$\frac{1}{\sqrt{2}}\left[x-\frac{\sin 2 x}{2}\right]$$ + C
= $$\frac{x}{\sqrt{2}}$$ – $$\frac{1}{2 \sqrt{2}}$$ sin 2x + C.

Question 17.
(i) ∫ cos 3x cos 5x dx
(ii) ∫ sin 4x sin 8x dx
(iii) ∫ sin 4x cos 3x dx
(iv) ∫ sin 2x cos 3x dx (NCERT)
Solution:
(i) ∫ cos 3x cos 5x dx
= $$\frac{1}{2}$$ ∫ (2 cos 5x cos 3x) dx
= $$\frac{1}{2}$$ ∫ [cos 8x + cos 2x] dx
= $$\frac{1}{2}\left[\frac{\sin 8 x}{8}+\frac{\sin 2 x}{2}\right]$$ + C

(ii) ∫ sin 4x sin 8x dx
= $$\frac{1}{2}$$ ∫ (2 sin 8x sin 4x) dx
= $$\frac{1}{2}$$ ∫ [cos 4x – cos 12 x] dx
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= $$\frac{1}{2}\left[\frac{\sin 4 x}{7}-\frac{\sin 12 x}{12}\right]$$ + C

(iii) ∫ sin 4x cos 3x dx
= $$\frac{1}{2}$$ ∫ 2 sin 4x cos 3x dx
= $$\frac{1}{2}$$ ∫ (sin 7x + sin x) dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= $$\frac{1}{2}\left[-\frac{\cos 7 x}{7}-\cos x\right]$$ + C
= $$\frac{1}{14}$$ [- cos 7x – 7 cos x] + C

(iv) ∫ sin 2x cos 3x dx
= $$\frac{1}{2}$$ ∫ (2 sin 2x cos 3x) dx
= $$\frac{1}{2}$$ ∫ [sin 5x + sin (- x)] dx
= $$\frac{1}{2}$$ ∫ (sin 5x – sin x) dx
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]
= $$\frac{1}{2}$$ [$$\frac{-\cos 5 x}{5}$$ + cos x] + C

Question 18.
(i) ∫ $$\frac{\sin 4 x}{\cos x}$$ dx
(ii) ∫ sin x sin 2x sin 3x dx
Solution:
(i) ∫ $$\frac{\sin 4 x}{\cos x}$$ dx
= ∫ $$\frac{2 \sin 2 x \cos 2 x}{\cos x}$$ dx
= ∫ $$\frac{2 \times 2 \sin x \cos x \cos 2 x d x}{\cos x}$$ dx
= 2 ∫ 2 cos 2x sin x dx
= 2 ∫ [sin 3x – sin x] dx
= 2 [- $$\frac{\cos 3 x}{3}$$ + cos x]
= $$\frac{2}{3}$$ (3 cos x – cos 3x)

(ii) ∫ sin x sin 2x sin 3x dx
= $$\frac{1}{2}$$ ∫ (2 sin 3x sin 2x) sin x dx
= $$\frac{1}{2}$$ ∫ [cos x – cos 5x] sin x dx
= $$\frac{1}{4}$$ ∫ 2 sin x cos x dx – $$\frac{1}{4}$$ ∫ 2 cos 5x sin x dx
= $$\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)$$ – $$\frac{1}{4}$$ ∫ [sin 6x – sin 4x] dx
= $$-\frac{\cos 2 x}{8}+\frac{\cos 6 x}{24}-\frac{\cos 4 x}{16}$$ + C

Question 19.
(i) ∫ sin6 x dx
(ii) ∫ sin4 x cos2 x dx.
Solution:
(i) ∫ sin6 x dx
= ∫ (sin2 x)3 dx
= ∫ $$\left[\frac{1-\cos 2 x}{2}\right]^3$$ dx
= $$\frac{1}{8}$$ ∫ [1 – $$\frac{1}{4}$$ (cos 6x + 3 cos 2x) – 3 cos 2x + 3 $$\left(\frac{1+\cos 4 x}{2}\right)$$]
[∵ cos3 θ = $$\frac{1}{4}$$ [cos 3θ + 3 cos θ]
and cos2 θ = $$\frac{1+\cos 2 \theta}{2}$$]
= $$\frac{1}{8 \times 4}$$ ∫ [4 – cos 6x – 3 cos 2x – 12 cos 2x + 6 + 6 cos 4x] dx
= $$\frac{1}{32}$$ ∫ [10 + 6 cos 4x – 15 cos 2x – cos 6x]
= $$\frac{1}{32}\left[10 x+\frac{6 \sin 4 x}{4}-\frac{15 \sin 2 x}{2}-\frac{\sin 6 x}{6}\right]$$ + C
= $$\frac{1}{192}$$ [60x + 9 sin 4x – 45 sin 2x – sin 6x] + C

(ii) ∫ sin4 x cos2 x dx.
= $$\frac{1}{4}$$ ∫ sin2 x (4 sin2 x cos2 x) dx
= $$\frac{1}{4}$$ ∫ $$\left[\frac{1-\cos 2 x}{2}\right]$$ [2 sin x cos x]2 dx
= $$\frac{1}{4} \int\left[\frac{1-\cos 2 x}{2}\right]\left[\frac{1-\cos 4 x}{2}\right]$$ dx
= $$\frac{1}{16}$$ ∫ [1 – cos 4x – cos 2x + (2 cos 4x cos 2x) $$\frac{1}{2}$$] dx
= $$\frac{1}{16}$$ ∫ [1 – cos 4x – cos 2x – $$\frac{1}{2}$$ (cos 6x + cos 2x)] dx
= $$\frac{1}{32}$$ ∫ [2 – 2 cos 4x – cos 2x + cos 6x] dx
= $$\frac{1}{32}\left[2 x-\frac{2 \sin 4 x}{4}-\frac{\sin 2 x}{2}+\frac{\sin 6 x}{6}\right]$$ + C
= $$\frac{1}{192}$$ [12x – 3 sin 4x – 3 sin 2x + sin 6x] + C

Question 20.
Evaluate ∫ cos mx cos nx dx, where m, n are positive integers, m ≠ n. What happens if m n?
Solution:
Case – I:
When m ≠ n
∴ ∫ cos mx cos nx dx = ∫ 2 cos mx cos nx dx
= $$\frac{1}{2}$$ ∫ [cos (m + n) x + cos (m – n) dx
= $$\frac{1}{2}\left[\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n) x}{m-n}\right]$$ + C

Case-II:
When m = n
∫ cos mx cos nx dx = ∫ cos2 mx dx
= ∫ $$\left[\frac{1+\cos 2 m x}{2}\right]$$ dx
= $$\frac{1}{2}\left[x+\frac{\sin 2 m x}{2 m}\right]$$ + C