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

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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) 4e^3x + 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) ∫ 4e^3x + 1 dx
= 4 ∫ e^3x 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) e^2x (NCERT)
Solution:
(i) ∫ sin mx dx = – \(\frac{\cos m x}{m}\) + C

(ii) ∫ e^2x dx
= \(\frac{e^{2 x}}{2}\) + C
[∵ ∫ e^mx 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
= ∫ e^x/2 dx
= \(\frac{e^{x / 2}}{1 / 2}\) + C
= 2e^x/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) ∫ e^{2x + 3} dx
(ii) ∫ e^{(3a log x + e3x log a)} dx
Solution:
(i) ∫ e^{2x + 3} dx
= \(\frac{e^{2 x+3}}{2}\) + C
[∵ ∫ e^mx dx = \(\frac{e^{m x}}{m}\) + C]

(ii) ∫ e^{(3a log x + e3x log a)} dx
= ∫ (e^{log x3a} + e^{log a3x}) dx
[∵ a log b = log b^a
and e^{log x} = x]
= ∫ (x^3a + 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) ∫ a^{3x + 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) ∫ a^{3x + 5} dx, a > 0
= \(\frac{a^{3 x+5}}{3(\log a)}\) + C
[∵ ∫ a^mx 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