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

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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 e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C, then the value of k is,
(a) 4
(b) – 2
(c) 2
(d) 1
Solution:
(c) 2

Given ∫ x e^kx2 dx = \(\frac{1}{4}\) e^2x2 + C
put x² = t
⇒ 2x dx = dt
⇒ ∫ e^kt \(\frac{dt}{2}\) = \(\frac{1}{4}\) e^2x2 + C
⇒ \(\frac{1}{2} \frac{e^{k t}}{k}\) = \(\frac{1}{4}\) e^2x2 + C
⇒ \(\frac{1}{2 k}\) e^2x2 + C = \(\frac{1}{4}\) e^2x2 + 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 (2^x) + 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 (2^x) + C
put 2^x = t
⇒ 2^x log 2 dx = dt
⇒ ∫ \(\frac{d t}{\log 2 \sqrt{1-t^2}}\) = k sin^-1 (2^x) + C
⇒ \(\frac{1}{\log 2}\) sin^-1 + C = k sin^-1(2^x + C
⇒ \(\frac{1}{\log 2}\) sin^-1 + C = k sin^-1 2^x + 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.
∫ \(\left(\frac{1-x}{1+x^2}\right)^2\) e^x 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.
∫ e^x (1 – cot x + cot² x) dx is equal to
(a) e^x cosec x + C
(b) – e^x cosec x + C
(c) e^x cot x + C
(d) – e^x cot x + C
Solution:
(d) – e^x cot x + C

Let I = ∫ e^x (1 – cot x + cot² x) dx
= ∫ e^x (1 – cot x + cosec² x – 1) dx
= ∫ e^x (cosec² x) dx – ∫ e^x cot x dx
= e^x (- cot x) – ∫ e^x (- cot x) dx
= e^x (- cot x) – ∫ e^x (- cot x) dx – ∫ e^x cot x dx + C
= – e^x 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.
∫ e^{3 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

∫ e^{3 log x} (x⁴ + 1)^-1 dx
= ∫ e^{log 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.
∫ e^{x log a} e^x 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

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

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

Question 31.
∫ \(\left(\frac{1+x+x^2}{1+x^2}\right)\) e^{tan-1 x} dx is equal to
(a) x + e^{tan-1 x} + C
(b) e^{tan-1 x} – x + C
(c) e^{tan-1 x} + C
(d) x e^{tan-1 x} + C
Solution:
(d) x e^{tan-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) – \(\frac{\pi}{4}\)
Answer:
(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 e^x = t
⇒ e^x 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 = \(\frac{\pi}{4}\)