ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.12

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 8 Integrals Ex 8.12

Very Short answer type questions (1 to 5) :

Evaluate the following (1 to 12) integrals

Question 1.
(i) ∫ e^x (sin x + cos x) dx (NCERT)
(ii) ∫ e^x (tan x + sec² x) dx
Solution:
(i) ∫ e^x (sin x + cos x) dx
= ∫ e^x sin x dx + ∫ e^x cos x dx
= sin x e^x – ∫ cos x e^x dx + ∫ e^x cos x dx + C
= sin x . e^x + C

(ii) Let I = ∫ e^x (tan x + sec² x) dx
= ∫ e^x tan x dx + ∫ e^x sec² x dx
= tan x . e^x – ∫ sec² x . e^x dx + ∫ e^x . sec² x dx + C
= tan x . e^x + C

Question 2.
(i) ∫ e^x (log x + \(\frac{1}{x}\)) dx
(ii) ∫ e^x (cot x – cosec² x) dx
Solution:
(i) Let I = ∫ e^x (log x + \(\frac{1}{x}\)) dx
= ∫ e^x log x dx + ∫ \(\frac{e^x}{x}\) dx
= log x e^x – ∫ \(\frac{1}{x}\) e^x dx + ∫ \(\frac{e^x}{x}\) dx + c
= e^x log x + c

(ii) Let I = ∫ e^x (cot x – cosec² x) dx
= ∫ e^x [cot x + (- cosec² x)] dx
= ∫ e^x cot x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ cosec² x e^x dx – ∫ e^x cosec² x dx
= e^x cot x + c

Question 3.
(i) ∫ e^x (x² + 2x) dx
(ii) ∫ (x + 1) e^x dx
Solution:
(i) Let I = ∫ e^x (x² + 2x) dx
= ∫ e^x x² dx + ∫ e^x . 2x dx
= x² e^x – 2x e^x dx + 2x . e^x dx + C
= x² e^x + C’

(ii) Let I = ∫ (x + 1) e^x dx
= ∫ x e^x dx + ∫ e^x dx
= x e^x – ∫ 1 . e^x dx + ∫ e^x + C
= x e^x + C

Question 4.
(i) ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx
(ii) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ e^x \(\left(\frac{1}{x}-\frac{1}{x^2}\right)\) dx

(ii) Let I = ∫ (1 + log x) dx
= ∫ 1 . dx + ∫ log x . 1 dx
= x + log x . x – ∫ \(\frac{1}{x}\) . x dx
= x + x log x – x + C
= x log x + C

Question 5.
(i) ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
(ii) ∫ (x cos x + sin x) dx
Solution:
(i) Let I = ∫ e^x (sin^-1 x + \(\frac{1}{\sqrt{1-x^2}}\)) dx
= ∫ e^x sin^-1 x dx + ∫ \(\frac{e^x}{\sqrt{1-x^2}}\) dx
= e^x sin^-1 x – ∫ \(\frac{1}{\sqrt{1-x^2}}\) e^x dx + ∫ \(\frac{e^x d x}{\sqrt{1-x^2}}\) + c
= e^x sin^-1 x + c

(ii) Let I = ∫ (x cos x + sin x) dx + ∫ sin x dx
= x sin x – ∫ sin x dx + ∫ sin x dx + C
= x sin x + C

Question 6.
(i) Given ∫ e^x (tan x + 1) sec x dx = e^x f(x) + C. Write f(x) satisfying the above.
(ii) If ∫ \(\left(\frac{x-1}{x^2}\right)\) e^x dx = f(x) e^x + C, then write the value of f(x).
Solution:
(i) Let I = ∫ e^x (tan x + 1) sec x dx
= ∫ e^x tan x sec x dx + ∫ e^x sec x dx
= e^x secx – ∫ e^x sec x dx + ∫ e^x sec x dx + C
= e^x sec x + C ……………..(1)
Also I = e^x f(x) + c ……………….(2)
From (1) and (2) ; we have
f(x) = sec x

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

Also given I = f(x) e^x + C …………(2)
From (1) and (2) ; we have
f(x) = \(\frac{1}{x}\)

Question 7.
(i) ∫ e^x (tan x + log sec x) dx
(ii) ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx
Solution:
(i) Let I = ∫ e^x (tan x + log sec x) dx
= ∫ e^x [tan x + (- log cos x)] dx
= – [ (log cos x) e^x – ∫ \(\frac{1}{cos x}\) (- sin x) e^x dx] + ∫ e^x tan x dx
= – e^x log cos x + c
= e^x log (cos x)^-1 + c
= e^x log sec x + c

(ii) Let I = ∫ \(\frac{2 x-1}{4 x^2}\) e^2x dx

Question 8.
(i) ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
(ii) ∫ e^x (cot x + log |sin x|) dx
Solution:
(i) Let I = ∫ \(\frac{\sin x \cos x-1}{\sin ^2 x}\) e^x dx
= ∫ e^x [cot x – cosec² x] dx
= ∫ e^x cot x – ∫ e^x cosec² x dx
= cot x e^x – ∫ (- cosec² x) e^x dx – ∫ e^x cosec² x dx
= e^x cot x + ∫ e^x cosec² x – ∫ e^x cosec² x dx + c
= e^x cot x + c

(ii) Let I = ∫ e^x (cot x + log |sin x|) dx
= ∫ e^x log sin x dx + ∫ e^x cot x dx
= (log sin x) e^x – ∫ \(\frac{1}{sin x}\) cos x e^x dx + ∫ e^x cot x dx
= e^x log sin x – ∫ cot x e^x dx + ∫ e^x cot x dx + C
= e^x log sin x + c

Question 9.
(i) ∫ \(\frac{x e^x}{(x+1)^2}\) dx (NCERT)
(ii) ∫ \(\frac{x-1}{(x+1)^3}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{x e^x}{(x+1)^2}\) dx

(ii) Let I = ∫ e^x \(\frac{x-1}{(x+1)^3}\) dx

Question 10.
(i) ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx (NCERT)
(ii) ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx
Solution:
(i) Let I = ∫ \(\frac{x-3}{(x-1)^3}\) e^x dx

(ii) Let I = ∫ \(\frac{x e^{2 x}}{(1+2 x)^2}\) dx

Question 11.
(i) ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx
(ii) ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
Solution:
(i) Let I = ∫ \(\frac{1+\sin x}{1+\cos x}\) e^x dx

(ii) Let I = ∫ \(\frac{2+\sin 2 x}{1+\cos 2 x}\) e^x dx
= ∫ \(\left[\frac{2+2 \sin x \cos x}{2 \cos ^2 x}\right]\) e^x dx
= ∫ sec² x . e^x dx + ∫ e^x . tan x dx
= e^x tan x – ∫ e^x . tan x dx + ∫ e^x tan x dx + C
= e^x tan x + C

Question 12.
(i) ∫ \(\frac{\sin 4 x-4}{1-\cos 4 x}\) e^x dx
(ii) ∫ (cos x + 3 sin x) e^3x dx
Solution:
(i) Let I = ∫ e^x \(\frac{\sin 4 x-4}{1-\cos 4 x}\) dx
= ∫ e^x \(\left(\frac{\sin 4 x-4}{2 \sin ^2 2 x}\right)\) dx
= ∫ e^x \(\left[\frac{2 \sin 2 x \cos 2 x-4}{2 \sin ^2 d x}\right]\) dx
[Form ∫ e^x [f(x) + f'(x)] dx]
= ∫ e^x [cot 2x – 2 cosec² 2x] dx
∴ I = ∫ e^x cot 2x dx – 2 ∫ e^x cosec² 2x dx
= cot 2x e^x – ∫ – cosec² 2x . 2 e^x dx – 2 ∫ e^x cosec² 2x dx
= e^x . cot 2x + C

(ii) Let I = ∫ (cos x + 3 sin x) e^3x dx
= ∫ e^3x cos x dx + ∫ sin x e^3x dx
= e^3x sin x – ∫ 3 e^3x sin x dx + 3 ∫ sin x e^3x dx + C
= e^3x sin x + C

Question 12 (old).
(i) ∫ (1 + log x) dx
Solution:
(i) Let I = ∫ (1 + log x) dx
= ∫ log x . dx + ∫ dx
= x log x – ∫ \(\frac{1}{x}\) . x dx + ∫ dx + C
= x log x + C

Question 13.
∫ (sin (log x) + cos (log x)) dx
Solution:
Let I = ∫ (sin (log x) + cos (log x)) dx
put log x = t
⇒ x = e^t
⇒ dx = e^t dt
∴ I = ∫ [sin t + cos t] e^t dt
= ∫ e^t sin t dt + ∫ e^t cos t dt
= sin t e^t – ∫ cos t e^t dt + ∫ e^t cos t dt
= e^t sin t + C
= x sin (log x) + C