Practice MCQs for JEE Mathematics Indefinite Integration
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Multiple Choice Questions
(a) π − x
(b) x − π
(c) \( \frac\pi2-x \)
(d) x
Show Answer & Explanation
Answer: (d) x
Explanation:
1. \( \cot^4x=\cot^2x\csc^2x-\csc^2x+1 \), so \( \int\cot^4x\,dx=-\frac{\cot^3x}3+\cot x+x+c \).
2. Hence φ(x) = x + c. Since φ(π/2) = π/2, c = 0 and φ(x) = x.
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(a) \( \frac23\frac{(1+x^2)^{3/2}}x+k \)
(b) \( \frac23(1+x^2)^{3/2}+k \)
(c) \( \frac23x(1+x^2)^{3/2}+k \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \frac23(1+x^2)^{3/2}+k \)
Explanation:
1. Put t = x²: \( \int\sqrt{1+t}\,dt=\frac23(1+t)^{3/2}+k=\frac23(1+x^2)^{3/2}+k \).
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(a) \( 2\sqrt{x^2+2}+k \)
(b) \( \sqrt{x^2+2}+k \)
(c) \( \frac1{(x^2+2)^{3/2}}+k \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( 2\sqrt{x^2+2}+k \)
Explanation:
1. d(x² + 1) = d(x² + 2); with u = x² + 2: \( \int u^{-1/2}du=2\sqrt u+k=2\sqrt{x^2+2}+k \).
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(a) \( \frac18(x^2-1)+k \)
(b) \( \frac12x^2+k \)
(c) \( \frac12x+k \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \frac12x^2+k \)
Explanation:
1. Let \( x=\cos\theta \): \( \sqrt{\frac{1-x}{1+x}}=\tan\frac\theta2 \), so \( 2\tan^{-1}(\cdot)=\theta=\cos^{-1}x \).
2. The integrand is cos(cos⁻¹x) = x, so the integral is \( \frac{x^2}2+k \).
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(a) \( x^x\log x+k \)
(b) \( e^{x^x}+k \)
(c) \( x^x+k \)
(d) none of these
Show Answer & Explanation
Answer: (c) \( x^x+k \)
Explanation:
1. Let y = xˣ: ln y = x ln x, so dy = xˣ(1 + ln x)dx.
2. The integral is \( \int dy=x^x+k \).
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(a) \( \sqrt{\log_e(3y-2)} \)
(b) \( \sqrt[3]{\log_e(3y-2)} \)
(c) \( \sqrt[3]{\log_e(2-3y)} \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \sqrt[3]{\log_e(3y-2)} \)
Explanation:
1. \( y=\frac{e^{x^3}}3+c \). At (0, 1): 1 = 1/3 + c, so c = 2/3.
2. Then \( e^{x^3}=3y-2 \), giving \( x=\sqrt[3]{\log_e(3y-2)} \).
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(a) \( \tan^{-1}x^2+k \)
(b) \( \frac12\tan^{-1}x^2+k \)
(c) \( \log(1+x^4)+k \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \frac12\tan^{-1}x^2+k \)
Explanation:
1. Put t = x²: x dx = dt/2, so \( \frac12\int\frac{dt}{1+t^2}=\frac12\tan^{-1}x^2+k \).
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63. .
(a) \( \log_2e\cdot\sin^{-1}(2^x)+k \)
(b) \( \sin^{-1}(2^x)+k \)
(c) \( \cos^{-1}(2^x)+k \)
(d) \( \frac1{\log_e2}+k \)
Show Answer & Explanation
Answer: (a) \( \log_2e\cdot\sin^{-1}(2^x)+k \)
Explanation:
1. Let t = 2ˣ: dt = 2ˣ ln 2 dx, so \( \int\frac{dt}{\ln2\sqrt{1-t^2}}=\frac1{\ln2}\sin^{-1}t+k \).
2. Since 1/ln 2 = log₂e, the answer is log₂e · sin⁻¹(2ˣ) + k.
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(a) \( \log_ex+\log_e(1+x^2)+k \)
(b) \( \log_ex+\tan^{-1}x+k \)
(c) \( \log_ex+2\tan^{-1}x+k \)
(d) none of these
Show Answer & Explanation
Answer: (c) \( \log_ex+2\tan^{-1}x+k \)
Explanation:
1. Write \( \frac{1+2x+x^2}{x(1+x^2)}=\frac1x+\frac{2}{1+x^2} \) (since 1 + x² over x(1 + x²) gives 1/x, and 2x over x(1 + x²) gives 2/(1 + x²)).
2. Integrating gives log x + 2 tan⁻¹x + k.
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(a) \( \frac27\log\left(x^{7/2}+\sqrt{x^7+1}\right)+c \)
(b) \( \frac12\log\frac{x^7+1}{x^7-1}+c \)
(c) \( 2\sqrt{1+x^7}+c \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( \frac27\log\left(x^{7/2}+\sqrt{x^7+1}\right)+c \)
Explanation:
1. Put u = x^{7/2}: du = (7/2) x^{5/2} dx, so \( \frac27\int\frac{du}{\sqrt{1+u^2}} \).
2. This equals \( \frac27\log(u+\sqrt{1+u^2})+c \).
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(a) \( \sqrt{1+x^{4/5}}+k \)
(b) \( \frac52\sqrt{1+x^{4/5}}+k \)
(c) \( x^{4/5}(1+x^{4/5})^{1/2}+k \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \frac52\sqrt{1+x^{4/5}}+k \)
Explanation:
1. Put u = x^{4/5}: du = (4/5) x^{−1/5} dx, so the integral is \( \frac54\int\frac{du}{\sqrt{1+u}}=\frac54\cdot2\sqrt{1+u}+k \).
2. This equals \( \frac52\sqrt{1+x^{4/5}}+k \).
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73. .
(a) cos x + x sin x
(b) −cos x − x sin x
(c) x sin x − cos x
(d) none of these
Show Answer & Explanation
Answer: (b) −cos x − x sin x
Explanation:
1. For π/2 < x < π, cos x < 0, so |cos x| = −cos x and the function is −x cos x.
2. By parts: \( -\int x\cos x\,dx=-(x\sin x+\cos x)=-x\sin x-\cos x \).
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67. .
(a) \( \frac12\log(\sec x^2+\tan x^2)+k \)
(b) \( \frac{x^2}2\log(\sec x^2+\tan x^2)+k \)
(c) \( 2\log(\sec x^2+\tan x^2)+k \)
(d) none of these
Show Answer & Explanation
Answer: (a) \( \frac12\log(\sec x^2+\tan x^2)+k \)
Explanation:
1. Put t = x²: x dx = dt/2.
2. Then \( \frac12\int\sec t\,dt=\frac12\log(\sec t+\tan t)+k \).
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69. .
(a) \( \log\frac{\varphi(x)}{f(x)}+k \)
(b) \( \frac12\left\{\log\frac{\varphi(x)}{f(x)}\right\}^2+k \)
(c) \( \frac{\varphi(x)}{f(x)}\log\frac{\varphi(x)}{f(x)}+k \)
(d) none of these
Show Answer & Explanation
Answer: (b) \( \frac12\left\{\log\frac{\varphi(x)}{f(x)}\right\}^2+k \)
Explanation:
1. Let \( t=\log\frac{\varphi}{f} \). Then \( dt=\left(\frac{\varphi'}{\varphi}-\frac{f'}f\right)dx=\frac{f\varphi'-f'\varphi}{f\varphi}dx \).
2. The integral is \( \int t\,dt=\frac{t^2}2+k=\frac12\left\{\log\frac\varphi f\right\}^2+k \).
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