Regular engagement with ML Aggarwal Class 12 Solutions ISC Chapter 8 Integrals MCQs can boost students’ confidence in the subject.

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

Choose the correct answer from the given four options in questions (1 to 55) :

Question 1.
∫ sin $$\frac{x}{2}$$ cos $$\frac{x}{2}$$ cos x dx is equal to
(a) – $$\frac{1}{4}$$ cos 2x + C
(b) – $$\frac{1}{8}$$ cos 2x + C
(c) $$\frac{1}{8}$$ cos 2x + C
(d) – $$\frac{1}{8}$$ cos 2x + C
Solution:
(b) – $$\frac{1}{8}$$ cos 2x + C

∫ sin $$\frac{x}{2}$$ cos $$\frac{x}{2}$$ cos x dx
= $$\frac{1}{2}$$ ∫ (2 sin $$\frac{x}{2}$$ cos $$\frac{x}{2}$$) cos x dx
= $$\frac{1}{2}$$ ∫ sin x cos x dx
= $$\frac{1}{4}$$ ∫ sin 2x dx
= $$\frac{1}{4}\left(-\frac{\cos 2 x}{2}\right)$$ + C
= – $$\frac{1}{8}$$ cos 2x + C

Question 2.
∫ $$\frac{d x}{\sin ^2 x \cos ^2 x}$$ is equal to
(a) tan x + cot x + C
(b) (tan x + cot x)2 + C
(c) tan x – cot x + C
(d) (tan x – cot x)2 + C
Solution:
∫ $$\frac{d x}{\sin ^2 x \cos ^2 x}$$
= ∫ $$\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x \cos ^2 x}$$ dx
= ∫ $$\left[\frac{1}{\cos ^2 x}+\frac{1}{\sin ^2 x}\right]$$ dx
= ∫ [sec2 x + cosec2 x] + C
= tan x – cot x + C

Question 3.
∫ $$\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$$ dx is equal to
(a) 2 (sin x + x cos α) + C
(b) 2 (sin x – x cos α) + C
(c) 2 (sin x + 2x cos α) + C
(d) 2 (sin x – 2x cos α) + C
Solution:
(a) 2 (sin x + x cos α) + C

∫ $$\frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$$ dx
= 2 ∫ $$\frac{2 \cos ^2 x-1-\left(2 \cos ^2 \alpha-1\right)}{\cos x-\cos \alpha}$$ dx
= 2 ∫ $$\frac{(\cos x-\cos \alpha)(\cos x+\cos \alpha)}{\cos x-\cos \alpha}$$ dx
= 2 [sin x + cos x] + C

Question 4.
∫ $$\frac{x}{4+x^4}$$ dx is equal to
(a) $$\frac{1}{4}$$ tan-1 x2 + C
(b) $$\frac{1}{2}$$ tan-1 $$\frac{x^2}{2}$$ + C
(c) $$\frac{1}{4}$$ tan-1 $$\left(\frac{x^2}{2}\right)$$ + C
(d) $$\frac{1}{2}$$ tan-1 2x2 + C
Solution:
(c) $$\frac{1}{4}$$ tan-1 $$\left(\frac{x^2}{2}\right)$$ + C

∫ $$\frac{x}{4+x^4}$$ dx ;
put x2 = t
⇒ 2x dx = dt

Question 5.
∫ $$\frac{\cos \sqrt{x}}{\sqrt{x}}$$ is equal to
(a) 2 cos √x + C
(b) 2 sin √x + C
(c) – 2 sin √x + C
(d) sin √x + C
Solution:
(b) 2 sin √x + C

put √x = t
⇒ $$\frac{1}{2 \sqrt{x}}$$ dx = dt
∴ ∫ $$\frac{\cos \sqrt{x}}{\sqrt{x}}$$ dx = ∫ cos t . 2t dt
= 2 sin t + C
= 2 sin √x + C

Question 6.
If ∫ x ekx2 dx = $$\frac{1}{4}$$ e2x2 + C, then the value of k is,
(a) 4
(b) – 2
(c) 2
(d) 1
Solution:
(c) 2

Given ∫ x ekx2 dx = $$\frac{1}{4}$$ e2x2 + C
put x2 = t
⇒ 2x dx = dt
⇒ ∫ ekt $$\frac{dt}{2}$$ = $$\frac{1}{4}$$ e2x2 + C
⇒ $$\frac{1}{2} \frac{e^{k t}}{k}$$ = $$\frac{1}{4}$$ e2x2 + C
⇒ $$\frac{1}{2 k}$$ e2x2 + C = $$\frac{1}{4}$$ e2x2 + C
∴ 2k = 4
⇒ k = 2

Question 7.
If ∫ x6 sin (5x7) dx = $$\frac{k}{5}$$ cos (5x7) + C, then the value of k is
(a) – $$\frac{1}{7}$$
(b) $$\frac{1}{7}$$
(c) $$\frac{1}{35}$$
(d) – $$\frac{1}{35}$$
Solution:
(a) – $$\frac{1}{7}$$

∫ x6 sin (5x7) dx = $$\frac{k}{5}$$ cos (5x7) + C
put x7 = t
⇒ 7x6 dx = dt

Question 8.
If ∫ $$\frac{2^x}{\sqrt{1-4^x}}$$ dx = k sin-1 (2x) + C, then the value of k is
(a) log 2
(b) $$\frac{1}{\log 2}$$
(c) $$\frac{2}{\log 2}$$
(d) $$\frac{\log 2}{2}$$
Solution:
(b) $$\frac{1}{\log 2}$$

Given ∫ $$\frac{2^x}{\sqrt{1-4^x}}$$ dx = k sin-1 (2x) + C
put 2x = t
⇒ 2x log 2 dx = dt
⇒ ∫ $$\frac{d t}{\log 2 \sqrt{1-t^2}}$$ = k sin-1 (2x) + C
⇒ $$\frac{1}{\log 2}$$ sin-1 + C = k sin-1(2x + C
⇒ $$\frac{1}{\log 2}$$ sin-1 + C = k sin-1 2x + C
∴ k = $$\frac{1}{\log 2}$$

Question 9.
If ∫ |x| dx = kx |x| + C, x ≠ 0, then the value of k is
(a) 2
(b) – 2
(c) – $$\frac{1}{2}$$
(d) $$\frac{1}{2}$$
Solution:
(d) $$\frac{1}{2}$$

Let I = ∫ |x| . 1 dx
= |x| . x – ∫ $$\frac{x}{|x|}$$ . x dx
= x |x| – ∫ $$\frac{|x|^2}{|x|}$$ dx
[∵ x2 = |x|2]
= x |x| – ∫ |x| dx
[∵ x ≠ 0]
⇒ I = x |x| – I
⇒ 2I = x |x|
⇒ I = $$\frac{x|x|}{2}$$ + C
Now I = kx |x| + C
⇒ k = $$\frac{1}{2}$$.

Question 10.
∫ cot x log (sin x) dx is equal to
(a) $$\frac{1}{2}$$ (log (sin x)2 + C
(b) log (sin x) + C
(c) $$\frac{1}{2}$$ (log (cos x))2 + C
(d) none of these
Solution:
(a) $$\frac{1}{2}$$ (log (sin x)2 + C

Let I = ∫ cot x log (sin x) dx
put log sin x = t
⇒ $$\frac{1}{\sin x}$$ cos x dx = dt
⇒ cot x dx = dt
= ∫ t dt
= $$\frac{t^2}{2}$$ + C
= $$\frac{1}{2}$$ [log (sin x)]2 + C

Question 11.
∫ $$\frac{x+\sin x}{1+\cos x}$$ dx is equal to
(a) log |1 + cos x| + C
(b) log |x + sin x| + C
(c) x – tan $$\frac{x}{2}$$ + C
(d) x tan $$\frac{x}{2}$$ + C
Solution:
(d) x tan $$\frac{x}{2}$$ + C

Question 12.
∫ $$\left(\frac{1-x}{1+x^2}\right)^2$$ ex dx is equal to
(a) $$\frac{e^x}{1+x^2}$$ + C
(b) – $$\frac{e^x}{1+x^2}$$ + C
(c) $$\frac{e^x}{\left(1+x^2\right)^2}$$ + C
(d) – $$\frac{e^x}{\left(1+x^2\right)^2}$$ + C
Solution:
(a) $$\frac{e^x}{1+x^2}$$ + C

Question 13.
∫ (x – 1)e-x dx is equal to
(a) (x – 2) e-x + C
(b) xe-x + C
(c) – xe-x + C
(d) (x + 1)e-x + C
Solution:
(c) – xe-x + C

∫ (x – 1)e-x dx = (x – 1) $$\frac{e^{-x}}{-1}$$ – ∫ 1 . $$\frac{e^{-x}}{-1}$$ dx +C
= – (x – 1) e-x – e-x + C
= – (x – 1 + 1) e-x + C
= – x e-x + C

Question 14.
∫ ex (1 – cot x + cot2 x) dx is equal to
(a) ex cosec x + C
(b) – ex cosec x + C
(c) ex cot x + C
(d) – ex cot x + C
Solution:
(d) – ex cot x + C

Let I = ∫ ex (1 – cot x + cot2 x) dx
= ∫ ex (1 – cot x + cosec2 x – 1) dx
= ∫ ex (cosec2 x) dx – ∫ ex cot x dx
= ex (- cot x) – ∫ ex (- cot x) dx
= ex (- cot x) – ∫ ex (- cot x) dx – ∫ ex cot x dx + C
= – ex cot x + C

Question 15.
If ∫ $$\frac{1+\cos 4 x}{\cot x-\tan x}$$ dx = k cos 4x + C, then the value of k is
(a) $$\frac{1}{4}$$
(b) – $$\frac{1}{2}$$
(c) – $$\frac{1}{8}$$
(d) – $$\frac{1}{4}$$
Solution:
(c) – $$\frac{1}{8}$$

Let I = ∫ $$\frac{1+\cos 4 x}{\cot x-\tan x}$$ dx
= ∫ $$\frac{2 \cos ^2 2 x}{\left(\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}\right)}$$ dx
= ∫ $$\frac{\frac{2 \cos ^2 2 x d x}{\cos ^2 x-\sin ^2 x}}{\sin x \cos x}$$ dx
= ∫ $$\frac{2 \cos ^2 2 x d x}{\frac{\cos 2 x}{\sin x \cos x}}$$ dx
= ∫ cos 2x (2 sin x cos x) dx
= ∫ cos 2x sin 2x dx
= $$\frac{1}{2}$$ ∫ sin 4x dx
= – $$\frac{1}{2} \frac{\cos 4 x}{4}$$ + C
= – $$\frac{1}{8}$$ cos 4x + C
Also, I = k cos 4x + C
∴ k = – $$\frac{1}{8}$$

Question 16.
∫ $$\frac{d x}{e^x+e^{-x}+2}$$ is equal to
(a) $$\frac{1}{e^x+1}$$ + C
(b) $$\frac{1}{1+e^{-x}}$$ + C
(c) – $$\frac{1}{e^x+1}$$ + C
(d) none of these
Solution:
(c) – $$\frac{1}{e^x+1}$$ + C

Question 17.
∫ $$\frac{(\log x)^5}{x}$$ dx is equal to
(a) $$\frac{\log x^6}{6}$$ + C
(b) $$\frac{(\log x)^6}{6}$$ + C
(c) $$\frac{(\log x)^6}{3 x^2}$$ + C
(d) none of these
Solution:
(b) $$\frac{(\log x)^6}{6}$$ + C

∫ $$\frac{(\log x)^5}{x}$$ dx ;
put log x = t
⇒ $$\frac{1}{x}$$ dx = dt
= ∫ t5 dt
= $$\frac{t^6}{6}$$ + C
= $$\frac{(\log x)^6}{6}$$ + C

Question 18.
∫ $$\frac{d x}{\sqrt{2 x-x^2}}$$ is equal to
(a) sin-1 (x – 1) + C
(b) sin-1 (x – 1) + C
(c) – $$\sqrt{2 x-x^2}$$ + C
(d) none of these
Solution:
(a) sin-1 (x – 1) + C

Question 19.
∫ $$\frac{x^2+1}{x^2-1}$$ dx is equal to
(a) x + log $$\left|\frac{x+1}{x-1}\right|$$ + C
(b) x + log $$\left|\frac{x-1}{x+1}\right|$$ + C
(c) log |(x – 1) (x + 1)| + C
(d) log |x2 + 1| + C
Solution:
(b) x + log $$\left|\frac{x-1}{x+1}\right|$$ + C

Question 20.
∫ (sin6 x + cos6 x + 3 sin2 x cos2 x) dx is equal to
(a) x + C
(b) $$\frac{3}{2}$$ sin 2x + C
(c) – $$\frac{3}{2}$$ cos 2x + C ‘
(d) none of these
Solution:
(a) x + C

∫ (sin6 x + cos6 x + 3 sin2 x cos2 x) dx
= ∫ [(sin2 x)3 + (cos2 x)3 + 3 sin2 x cos2 x (sin2 x + cos2 x)] dx
= ∫ (sin2 x+ cos2 x)3 dx
= ∫ dx
= x + C

Question 21.
∫ $$\frac{d x}{x\left(x^7+1\right)}$$ is equal to
(a) log $$\left|\frac{x^7}{x^7+1}\right|$$ + C
(b) $$\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|$$ + C
(c) log $$\left|\frac{x^7+1}{x^7}\right|$$ + C
(d) $$\frac{1}{7} \log \left|\frac{x^7+1}{x^7}\right|$$ + C
Solution:
(b) $$\frac{1}{7} \log \left|\frac{x^7}{x^7+1}\right|$$ + C

Let I = ∫ $$\frac{d x}{x\left(x^7+1\right)}$$ ;
put x7 = t
⇒ 7x6 dx = dt

Question 22.
∫ (sin4 x – cos4 x) dx is equal to
(a) $$\frac{1}{2}$$ cos 2x + C
(b) – $$\frac{1}{2}$$ cos 2x + C
(c) $$\frac{1}{2}$$ sin 2x + C
(d) – $$\frac{1}{2}$$ sin 2x + C
Solution:
(d) – $$\frac{1}{2}$$ sin 2x + C

∫ (sin4 x – cos4 x) dx = ∫ (sin2 x – cos2 x) (sin2 x + cos2 x) dx
= – ∫ cos 2x dx
= – $$\frac{\sin 2 x}{2}$$ + C

Question 23.
∫ $$\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}$$ dx is equal to
(a) 3 (tan-1 x)2 + C
(b) (tan-1 x)4 + C
(c) (tan-1 x)4 + C
(d) none of these
Solution:
(b) (tan-1 x)4 + C

put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
∴ I = ∫ $$\frac{\left(\tan ^{-1} x\right)^3}{1+x^2}$$ dx
= ∫ t3 dt
= $$\frac{t^4}{4}$$ + C
= $$\frac{\left(\tan ^{-1} x\right)^4}{4}$$ + C

Question 24.
∫ e3 log x (x4 + 1)-1 dx is equal to
(a) $$\frac{1}{4}$$ log (x4 + 1) + C
(b) – $$\frac{1}{4}$$ (x4 + 1) + C
(c) log (x4 + 1) + C
(d) none of these
Solution:
(a) $$\frac{1}{4}$$ log (x4 + 1) + C

∫ e3 log x (x4 + 1)-1 dx
= ∫ elog x3 $$\frac{1}{x^4+1}$$ dx
= ∫ $$\frac{x^3 d x}{x^4+1}$$ ;
put x4 + 1 = t
⇒ 4x3 dx = dt
= $$\frac{1}{4} \int \frac{d t}{t}$$
= $$\frac{1}{4}$$ log |t| + C
= $$\frac{1}{4}$$ log |x4 + 1| + C

Question 25.
∫ (sin (log x) + cos (log x)) dx is equal to
(a) sin (log x) + C
(b) cos (log x) + C
(c) x cos (log x) + C
(d) x sin (log x) + C
Solution:
(d) x sin (log x) + C

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

Question 26.
∫ $$\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}$$ dx is equal to
(a) $$\frac{1}{2} \sqrt{1+x}$$ + C
(b) $$\frac{1}{2}$$ (1 + x)3/2 + C
(c) 2 (1 + x)3/2 + C
(d) $$\sqrt{1 + x}$$ + C
Solution:
(b) $$\frac{1}{2}$$ (1 + x)3/2 + C

Question 27.
∫ $$\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}$$ dx is equal to
(a) f(x) log (f(x)) + C
(b) log (log (f(x)) + C
(c) $$\frac{f(x)}{\log (f(x))}$$ + C
(d) $$\frac{1}{\log (\log (f(x)))}$$ + C
Solution:
(b) log (log (f(x)) + C

Let I = ∫ $$\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}$$ dx
put log (f(x)) = t
⇒ $$\frac{f^{\prime}(x)}{f(x)}$$ dx = dt
= ∫ $$\frac{1}{t}$$ dt
= |log |t| + C
= log |log f(x)| + C

Question 28.
∫ x3 log x dx is equal to
(a) $$\frac{x^4 \log x}{4}$$ + C
(b) $$\frac{x^4}{8}\left(\log x-\frac{4}{x^2}\right)$$ + C
(c) $$\frac{x^4}{16}$$ (4 log x – 1) + C
(d) $$\frac{x^4}{16}$$ (4 log x + 1) + C
Solution:
(c) $$\frac{x^4}{16}$$ (4 log x – 1) + C

∫ x3 log x dx

Question 29.
∫ ex log a ex dx is equal to
(a) $$\frac{a^x}{\log a e}$$ + C
(b) $$\frac{e^x}{1+\log a}$$
(c) (ae)x + C
(d) $$\frac{(a e)^x}{\log a e}$$ + C
Solution:
(d) $$\frac{(a e)^x}{\log a e}$$ + C

∫ ex log a ex dx = ∫ elog ax . ex dx
= ∫ ax . ex dx
= ∫ (ae)x dx
= $$\frac{(a e)^x}{\log a e}$$ + C

Question 30.
∫ $$\left(\frac{1-\sin x}{1-\cos x}\right)$$ ex dx is equal to
(a) – ex tan $$\frac{x}{2}$$ + C
(b) – ex cot $$\frac{x}{2}$$ + C
(c) – $$\frac{1}{2}$$ ex tan $$\frac{x}{2}$$ + C
(d) – $$\frac{1}{2}$$ ex cot $$\frac{x}{2}$$ + C
Solution:
(b) – ex cot $$\frac{x}{2}$$ + C

Question 31.
∫ $$\left(\frac{1+x+x^2}{1+x^2}\right)$$ etan-1 x dx is equal to
(a) x + etan-1 x + C
(b) etan-1 x – x + C
(c) etan-1 x + C
(d) x etan-1 x + C
Solution:
(d) x etan-1 x + C

Question 32.
If $$\int_0^{40} \frac{d x}{2 x+1}$$ = log k, then value of k is
(a) 3
(b) $$\frac{9}{2}$$
(c) 9
(d) none of these
Solution:
(c) 9

Given $$\int_0^{40} \frac{d x}{2 x+1}$$ = log k
⇒ $$\left.\frac{\log |2 x+1|}{2}\right]_0^{40}$$ = log k
⇒ $$\frac{1}{2}$$[log 81 – log 1] = log k
⇒ $$\frac{1}{2}$$ log 92 = log k
⇒ log 9 = log k
∴ k = 9

Question 33.
$$\int_0^{\sqrt{3}} \frac{d x}{1+x^2}$$ is equal to
(a) $$\frac{\pi}{3}$$
(b) $$\frac{\pi}{4}$$
(c) $$\frac{\pi}{6}$$
(d) $$\frac{\pi}{12}$$
Solution:
(d) $$\frac{\pi}{12}$$

$$\left.\int_0^{\sqrt{3}} \frac{d x}{1+x^2}=\tan ^{-1} x\right]_1^{\sqrt{3}}$$
= tan-1 √3 – tan-1 1
= $$\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$

Question 34.
If $$\int_0^k \frac{1}{9 x^2+1} d x=\frac{\pi}{12}$$, then k is equal to
(a) $$\frac{\pi}{4}$$
(b) $$\frac{1}{3}$$
(c) 3
(d) none of these
Solution:
(b) $$\frac{1}{3}$$

Question 35.
$$\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}$$ dx is equal to
(a) 3 – log 12
(b) 1 – log 2
(c) 1 + log 2
(d) none of these
Solution:
(b) 1 – log 2

$$\int_0^{\pi / 2} \frac{\sin x \cos x}{1+\sin x}$$ dx ;
put sin x = t
⇒ cos x dx = dt
When x = 0 ⇒ t = 0 ;
When x = $$\frac{\pi}{2}$$ ⇒ t = 1
= $$\int_0^1 \frac{t d t}{1+t}$$
= $$\int_0^1\left[\frac{1+t-1}{1+t}\right]$$ dt
= $$\int_0^1\left[1-\frac{1}{1+t}\right]$$ dt
= t – log |1 + t|
= 1 – log 2 – 0 – 0
= 1 – log 2

Question 36.
The value of $$\int_{\pi / 6}^{\pi / 3} \frac{1}{\sin 2 x}$$ dx is
(a) log 3
(b) $$\frac{1}{2}$$ log 2
(c) $$\frac{1}{2}$$ log 3
(d) none of these
Solution:
(c) $$\frac{1}{2}$$ log 3

Question 37.
$$\int_{-1}^0 \frac{d x}{x^2+2 x+2}$$ is equal to
(a) 0
(b) $$\frac{\pi}{2}$$
(c) $$\frac{\pi}{4}$$
(d) – $$\frac{\pi}{4}$$
(c) $$\frac{\pi}{4}$$

$$\int_{-1}^0 \frac{d x}{x^2+2 x+2}$$ = $$\int_{-1}^0 \frac{d x}{(x+1)^2+1^2}$$
= tan-1 x$$]_{-1}^0$$
= tan-1 1 – tan-1 0
= $$\frac{\pi}{4}$$ – 0 = $$\frac{\pi}{4}$$

Question 38.
$$\int_0^1 \frac{\tan ^{-1} x}{1+x^2}$$ dx is equal to
(a) $$\frac{\pi}{4}$$
(b) $$\frac{\pi^2}{32}$$
(c) 1
(d) $$\frac{\pi^2}{16}$$
Solution:
(b) $$\frac{\pi^2}{32}$$

$$\int_0^1 \frac{\tan ^{-1} x}{1+x^2}$$ dx ;
put tan-1 x = t
⇒ $$\frac{1}{1+x^2}$$ dx = dt
When x = 0
⇒ t = tan-1 0 = 0
When x = 1
⇒ t = tan-1 1 = $$\frac{\pi}{4}$$
= $$\left.\int_0^{\pi / 4} t d t=\frac{t^2}{2}\right]_0^{\pi / 4}$$
= $$\frac{\pi^2}{32$$

Question 39.
$$\int_0^1 \frac{d x}{e^x+e^{-x}}$$ is equal to
(a) $$\frac{\pi}{4}$$
(b) $$\frac{\pi}{2}$$
(c) tan-1 $$\left(\frac{e-1}{e+1}\right)$$
(d) tan-1 $$\left(\frac{1-e}{1+e}\right)$$
Solution:
(c) tan-1 $$\left(\frac{e-1}{e+1}\right)$$

$$\int_0^1 \frac{d x}{e^x+e^{-x}}=\int_0^1 \frac{e^x d x}{e^{2 x}+1}$$ ;
put ex = t
⇒ ex dx = dt
When x = 0 ⇒ t = 1
When x = 1 ⇒ t = e
= $$\int_1^e \frac{d t}{t^2+1}$$
= tan-1 t$$]_1^e$$
= tan-1 e – tan-1 1
= tan-1 e – $$\frac{\pi}{4}$$
= tan-1 $$\left(\frac{e-1}{e+1}\right)$$

Question 40.
$$\int_0^1 \frac{x^4+1}{x^2+1}$$ dx is equal to
(a) $$\frac{1}{6}$$ (3 – 4π)
(b) $$\frac{1}{6}$$ (3π + 4)
(c) $$\frac{1}{6}$$ (3 + 4π)
(d) $$\frac{1}{6}$$ (3π – 4)
Solution:
(d) $$\frac{1}{6}$$ (3π – 4)

Question 41.
$$\int_{a+c}^{b+c}$$ f(x) dx is equal to
(a) $$\int_a^b\int_a^b$$ f(x – c) dx
(b) $$\int_a^b$$ f(x + c) dx
(c) $$\int_a^b$$ f(x) dx
(d) $$\int_{a-c}^{b-c}$$ f(x) dx
Solution:
(b) $$\int_a^b$$ f(x + c) dx

put x = t + c
⇒ dx = dt
When x = a + c
⇒ t = a
When x = b + c
⇒ t = b
∴ I = $$\int_a^b$$ f(t + c) dt
= $$\int_a^b$$ f(x + c) dx
[∵ $$\int_a^b$$ f(x) dx = $$\int_a^b$$ f(t) dt]

Question 42.
$$\int_{-\pi / 2}^{\pi / 2}$$ cos x dx is equal to
(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2

$$\int_{-\pi / 2}^{\pi / 2}$$ cos x dx = + sin x$$]_{-\pi / 2}^{\pi / 2}$$
= sin $$\frac{\pi}{2}$$ – sin (- $$\frac{\pi}{2}$$)
= 1 + sin $$\frac{\pi}{2}$$
= 1 + 1 = 2

Question 43.
$$\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}$$ is equla to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1

Question 44.
$$\int_0^{\pi / 2}$$ log |tan x| dx is equal to
(a) π – log 2
(b) – π log 2
(c) 0
(d) none of these
Solution:
(c) 0

Let I = $$\int_0^{\pi / 2}$$ log |tan x| dx
I = $$\int_0^{\pi / 2}$$ log tan x dx ………………(1)
[when 0 ≤ x < $$\frac{\pi}{2}$$ ⇒ tan x > 0
⇒ |tan x| = tan x]
We know that
$$\int_0^a$$ f(x) dx = $$\int_0^a$$ f(a – x) dx
∴ I = $$\int_0^{\pi / 2}$$ log tan ($$\frac{\pi}{2}$$ – x) dx
= $$\frac{\pi}{2}$$ log cot x dx ………………(2)
On adding (1) and (2) ; we have
2I = $$\int_0^{\pi / 2}$$ (log tan x + log cot x) dx
= $$\int_0^{\pi / 2}$$ log (tan x . cot x) dx
= $$\int_0^{\pi / 2}$$ log 1 dx = 0
⇒ I = 0

Question 45.
$$\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}$$ dx is equal to
(a) 0
(b) $$\frac{1}{2}$$
(c) 1
(d) 2
Solution:
(d) 2

$$\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{1-\cos 2 x}{2}}$$ dx
= $$\int_{-\pi / 2}^{\pi / 2} \sqrt{\frac{2 \sin ^2 x}{2}}$$
= $$\int_{-\pi / 2}^{\pi / 2}$$ |sin x| dx
= $$\int_{-\pi / 2}^0$$ |sin x| dx + $$\int_0^{\pi / 2}$$ |sin x| dx
When – $$\frac{\pi}{2}$$ ≤ x < 0
⇒ sin x ≥ 0
⇒ |sin x| = – sin x
When 0 ≤ x ≤ $$\frac{\pi}{2}$$
⇒ sin x ≥ 0
⇒ |sin x| = sin x
= $$\int_{-\pi / 2}^0$$ – sin x dx + $$\int_0^{\pi / 2}$$ sin x dx
= $$\left.\cos x]_{-\pi / 2}^0+(-\cos x)\right]_0^{\pi / 2}$$
= (1 – 0) + (- 0 + 1) = 2

Question 46.
$$\int_{-\pi}^\pi \frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}$$ dx is equal to
(a) 2π
(b) π
(c) $$\frac{\pi}{2}$$
(d) $$\frac{\pi}{4}$$
Solution:
(b) π

Question 47.
$$\int_a^b \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a+b-x}}$$ dx is equal to
(a) $$\frac{\pi}{2}$$
(b) π
(c) $$\frac{1}{2}$$ (b – a)
(d) b – a
Solution:
(c) $$\frac{1}{2}$$ (b – a)

Question 48.
$$\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}$$ dx is equal to
(a) 0
(b) $$\frac{\pi}{6}$$
(c) $$\frac{\pi}{4}$$
(d) $$\frac{\pi}{2}$$
Solution:
(a) 0

Let I = $$\int_0^{\pi / 2} \frac{\cos x-\sin x}{2+\sin x \cos x}$$ dx ……………..(1)
We know that
$$\int_0^a$$ f(x) dx = $$\int_0^a$$ f(a – x) dx
∴ I = $$\int_0^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{2+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}$$
I = $$\int_0^{\pi / 2} \frac{\sin x-\cos x}{2+\cos x \sin x}$$ ………………….(2)
On adding (1) and(2) ; we have
2I = $$\int_0^{\pi / 2} \frac{\cos x-\sin x+\sin x-\cos x}{2+\cos x \sin x}$$ dx = 0
⇒ I = 0

Question 49.
The value of $$\int_0^2\left|\cos \frac{\pi}{2} t\right|$$ dt is equal to
(a) $$\frac{3}{4 \pi}$$
(b) $$\frac{4}{\pi}$$
(c) $$\frac{\pi}{2}$$
(d) 2π
Solution:
(b) $$\frac{4}{\pi}$$

Since, 0 ≤ t ≤ 2
⇒ 0 ≤ $$\frac{\pi}{2}$$ t ≤ π
⇒ cos $$\frac{\pi}{2}$$ t > 0
and 1 ≤ t ≤ 2
⇒ $$\frac{\pi}{2}$$ ≤ $$\frac{\pi}{2}$$ t ≤ π
⇒ cos $$\frac{\pi}{2}$$ t < 0

Question 50.
$$\int_0^{\pi / 2} \sqrt{1-\sin 2 x}$$ dx is equal to
(a) 2√2
(b) 2 (√2 + 1)
(c) 2
(d) 2 (√2 – 1)
Solution:
(d) 2 (√2 – 1)

Let I = $$\int_0^{\pi / 2} \sqrt{1-\sin 2 x}$$ dx
= $$\int_0^{\pi / 2} \sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x}$$
= $$\int_0^{\pi / 2} \sqrt{(\sin x-\cos x)^2}$$ dx
= $$\int_0^{\pi / 2}$$ |sin x – cos x| dx
When 0 ≤ x < $$\frac{\pi}{4}$$ ⇒ cos x > sin x
⇒ sin x – cos x < 0
∴ |sin x – cos x| = – (sin x – cos x)
and $$\frac{\pi}{4}$$ ≤ x ≤ $$\frac{\pi}{2}$$
⇒ sin x > cos x
⇒ sin x – cos x > 0
|sin x – cos x| = sin x – cos x
= $$\int_0^{\pi / 4}$$ – (sin x -cos x) dx + $$\int_{\pi / 4}^{\pi / 2}$$ (sin x – cos x) dx
= – [- cos x – sin x$$]_0^{\pi / 4}$$ + (- cos x- sin x)$$]_{\pi / 4}^{\pi / 2}$$
= $$\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(1+0)\right]+\left(0-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)$$
= √2 – 1 – 1 + √2
= 2 (√2 – 1)

Question 51.
$$\int_0^{\pi / 2}$$ sin 2x log cot x dx is equal to
(a) π
(b) 2π
(c) 0
(d) $$\frac{\pi}{2}$$
Solution:
(c) 0

Let I = $$\int_0^{\pi / 2}$$ sin 2x log cot x dx ………..(1)
We know that
$$\int_0^a$$ f(x) dx = $$\int_0^a$$ f (a – x) dx
∴ I = $$\int_0^{\pi / 2}$$ sin 2 ($$\frac{\pi}{2}$$ – x) log cot ($$\frac{\pi}{2}$$ – x) dx
= $$\int_0^{\pi / 2}$$ sin (π – x) log cot ($$\frac{\pi}{2}$$ – x) dx
∴ I = $$\int_0^{\pi / 2}$$ sin x log tan x dx ……………..(2)
On adding (1) and (2) ; we have
2I = $$\int_0^{\pi / 2}$$ sin 2x [log cot x + log tan x] dx
= $$\int_0^{\pi / 2}$$ sin 2x log (cot x . tan x) dx
= $$\int_0^{\pi / 2}$$ sin 2x log 1 dx = 0
[∵ log 1 = 0 ]
⇒ I = 0

Question 52.
The value of $$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x sin2 x + 2 tan-1 x – 1) dx is
(a) π
(b) 2
(c) 1
(d) 0
Solution:
(a) π

Let I = $$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x sin2 x + 2 tan-1 x – 1) dx
= $$\int_{-\pi / 2}^{\pi / 2}$$ (x5 + x sin2 x + 2 tan-1 x) dx + $$\int_{-\pi / 2}^{\pi / 2}$$ dx
= I1 + x$$]_{-\pi / 2}^{\pi / 2}$$
= I1 + $$\left(\frac{\pi}{2}+\frac{\pi}{2}\right)$$
= I1 + π …………..(1)
Let f(x) = x5 + x sin2 x + 2 tan-1 x
∴ f(- x) = (- x)5 – x (sin (- x))2 + 2 tan-1 (- x)
= – x5 – x sin2 x – 2 tan-1 x
= – f(x)
Thus f(x) is an odd function.
∴ I1 = $$\int_{-\pi / 2}^{\pi / 2}$$ f(x) = 0
∴ from (1) ;
I = 0 + π = π

Question 53.
If f (a + b – x) = f(x), then $$\int_a^b$$ x f(x) dx is equal to
(a) $$\frac{a+b}{2} \int_a^b$$ f (b – x) dx
(b) $$\frac{a+b}{2} \int_a^b$$ f(a – x) dx
(c) $$\frac{b-a}{2} \int_a^b$$ f(x) dx
(d) $$\frac{a+b}{2} \int_a^b$$ f(x) dx

Solution:
(d) $$\frac{a+b}{2} \int_a^b$$ f(x) dx

Let I = $$\int_a^b$$ x f(x) dx ………………(1)
We know that
$$\int_a^b$$ f(x) dx = $$\int_a^b$$ f(a + b – x) dx
I = $$\int_a^b$$ (a + b – x) f(a + b – x) dx
I = $$\int_a^b$$ (a + b – x) f(x) dx ………………(2)
[∵ f(a + b – x) = f(x)]
On adding (1) and (2) ; we get
2I = $$\int_a^b$$ (a + b – x + x) f(x) dx
⇒ I = $$\frac{a+b}{2} \int_a^b$$ f(x) dx

Question 54.
If a is a real number such that $$\int_0^a$$ x dx ≤ a + 4, then
(a) 0 ≤ a ≤ 4
(b) – 2 ≤ a ≤ 0
(c) a ≤ – 2 or a ≥ 4
(d) – 2 ≤ a ≤ 4
Solution:
(d) – 2 ≤ a ≤ 4

$$\int_0^a$$ x dx ≤ a + 4
⇒ $$\left.\frac{x^2}{2}\right]_0^a$$ ≤ a + 4
⇒ $$\frac{a^2}{2}$$ ≤ a + 4
⇒ a2 ≤ 2a + 8
⇒ (a2 – 2a – 8) ≤ 0
⇒ (a + 2) (a – 2) ≤ 0
⇒ – 2 ≤ a ≤ 4
[if (x – a) (x – b) ≤ 0 and a < b. Then a ≤ x ≤ b]

Question 55.
$$\int_0^{\pi / 8}$$ tan2 2x dx is equal to
(a) $$\frac{4-\pi}{8}$$
(b) $$\frac{4+\pi}{8}$$
(c) $$\frac{4-\pi}{4}$$
(d) $$\frac{4-\pi}{2}$$
Solution:
(a) $$\frac{4-\pi}{8}$$

Let I = $$\int_0^{\pi / 8}$$ tan2 2x dx
put 2x = t
⇒ 2 dx = dt
When x = 0 ⇒ t = 0 ;
When x = $$\frac{\pi}{8}$$
⇒ t = $$\frac{\pi}{4}$$