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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Medium

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61 Medium
61 of 411
Analyzing +4 | -1
If In=∫(sin⁡x+cos⁡x)ndx(n≥2)I_n=\int(\sin x+\cos x)^n d x(n \geq 2), then nIn−2(n−1)In−2n I_n-2(n-1) I_{n-2} is equal to
View answer and explanation
Correct option: B
(sin⁡x+cos⁡x)n−1(sin⁡x−cos⁡x)(\sin x+\cos x)^{n-1}(\sin x-\cos x)
Option B ((sin⁡x+cos⁡x)n−1(sin⁡x−cos⁡x)\sin x+\cos x)^{n-1}(\sin x-\cos x)) is correct because after integration by parts and simplification, the expression matches this form.
62 Medium
62 of 411
Analyzing +4 | -1
∫  (x+1)3/2  [sin⁡−1  (log⁡  x)  +  cos⁡−1  (log⁡  x)]  dx\int \,\,{\left( {x + 1} \right)^{3/2}}\,\,[{\sin ^{ - 1}}\,\,(\log \,\,x)\,\, + \,\,{\cos ^{ - 1}}\,\,(\log \,\,x)]\,\,dx is equal to
View answer and explanation
Correct option: C
Does not exist
Option C (Does not exist) is correct because the domain of the inverse trigonometric functions requires the argument to be within [-1, 1], which is not satisfied for all x in the integral.
63 Medium
63 of 411
Analyzing +4 | -1
∫Sin⁡−1x−Cos⁡−1xSin⁡−1x+Cos⁡−1xdx=\int \frac{\operatorname{Sin}^{-1} \sqrt{x}-\operatorname{Cos}^{-1} \sqrt{x}}{\operatorname{Sin}^{-1} \sqrt{x}+\operatorname{Cos}^{-1} \sqrt{x}} d x=
View answer and explanation
Correct option: A
4π[x−x22+xSin⁡−1x−12Sin⁡−11−x]−x+c\frac{4}{\pi}\left[\frac{\sqrt{x-x^2}}{2}+x \operatorname{Sin}^{-1} \sqrt{x}-\frac{1}{2} \operatorname{Sin}^{-1} \sqrt{1-x}\right]-x+c
The correct answer is A (4π[x−x22+xSin⁡−1x−12Sin⁡−11−x]−x+c\frac{4}{\pi}\left[\frac{\sqrt{x-x^2}}{2}+x \operatorname{Sin}^{-1} \sqrt{x}-\frac{1}{2} \operatorname{Sin}^{-1} \sqrt{1-x}\right]-x+c) because after simplifying the integral using trigonometric identities and integration by parts, the result matches option A.
64 Medium
64 of 411
Analyzing +4 | -1
Let x2≠n  π−1,   n∈N{x^2} \ne n\;\pi - 1,\;\,n \in N. The indefinite integral of
x  2  sin⁡  (x2+  1)  −  sin⁡  2(x2+  1)2  sin⁡  (x2+  1)  +  sin⁡  2(x2+  1)x\,\,\sqrt {\frac{{2\,\,\sin \,\,\left( {{x^2} + \,\,1} \right)\,\, - \,\,\sin \,\,2\left( {{x^2} + \,\,1} \right)}}{{2\,\,\sin \,\,\left( {{x^2} + \,\,1} \right)\,\, + \,\,\sin \,\,2\left( {{x^2} + \,\,1} \right)}}}

is but for an arbitrary constant
View answer and explanation
Correct option: B
log⁡  ∣ 12  sec⁡  (x2+12) ∣\log \,\,\left| {\,\frac{1}{2}\,\,\sec \,\,\left( {\frac{{{x^2} + 1}}{2}} \right)\,} \right|
Option B (\log \,\left| {\\,\frac{1}{2}\\,\sec \\,\left( {\frac{{{{{x^2} + 1}}}}{2}} \right)\} \right|\log \,\left| {\\,\frac{1}{2}\\,\sec \\,\left( {\frac{{{{{x^2} + 1}}}}{2}} \right)\} \right|) is correct because it is the result of the indefinite integral after simplification and substitution.
65 Medium
65 of 411
Analyzing +4 | -1
∫cos⁡2θlog⁡e(cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)dθ=sin⁡2θ2×log⁡e(cos⁡θ+sin⁡θcos⁡θ−sin⁡θ)−12f(x)+c\int \cos 2 \theta \log _e\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right) d \theta=\frac{\sin 2 \theta}{2}\times \log _e\left(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right)-\frac{1}{2} f(x)+c where f(x)f(x) is
View answer and explanation
Correct option: B
log⁡e(sec⁡2θ)\log _e(\sec 2 \theta)
The correct answer is B (log⁡e(sec⁡2θ)\log _e(\sec 2 \theta)) because using integration by parts, the integral simplifies, and the remaining integral is a standard form that results in the logarithm of secant.
66 Medium
66 of 411
Application +4 | -1
∫sec⁡xdxcos⁡2x=\int \frac{\sec x d x}{\sqrt{\cos 2 x}}=
View answer and explanation
Correct option: A
sin⁡−1(tan⁡x)\sin ^{-1}(\tan x)
Option A is correct because the integral can be rewritten and solved using a trigonometric substitution, resulting in sin⁡−1(tan⁡x)\sin^{-1}(\tan x).
67 Medium
67 of 411
Analyzing +4 | -1
∫a+bsin⁡x(b+asin⁡x)2dx\int\frac{a+b \sin x}{(b+a \sin x)^2} d x equals
View answer and explanation
Correct option: A
−(acos⁡xb+asin⁡x)+c-\left(\frac{a \cos x}{b+a \sin x}\right)+c
Option A (−(acos⁡xb+asin⁡x)+c-\left(\frac{{a \cos x}}{{b+a \sin x}}\right)+c) is correct because the integral simplifies to this expression after algebraic manipulation and recognizing the derivative of a function.
68 Medium
68 of 411
Application +4 | -1
∫(x+2x+4)2exdx\int\left(\frac{x+2}{x+4}\right)^2 e^x d x is equal to
View answer and explanation
Correct option: A
ex(xx+4)+ce^x\left(\frac{x}{x+4}\right)+c
Option A (ex(xx+4)+ce^x\left(\frac{{x}}{{x+4}}\right)+c) is correct because the integral can be solved by simplifying the expression and applying the rule ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
69 Medium
69 of 411
Application +4 | -1
The integral ∫(x8−x2)dx(x12+3x6+1)tan⁡−1(x3+1x3)\int \frac{\left(x^8-x^2\right) d x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} is equal to
View answer and explanation
Correct option: A
13ln⁡∣(tan⁡−1(x3+1x3))∣+c\frac{1}{3} \ln \left|\left(\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right)\right|+c
Option A (13ln⁡∣(tan⁡−1(x3+1x3))∣+c\frac{1}{3} \ln |(\tan ^{-1}(x^3+\frac{1}{x^3}))|+c) is correct because it is the result of a u-substitution and simplification of the integral.
70 Medium
70 of 411
Application +4 | -1
∫dxcos⁡x+3sin⁡x\int \frac{d x}{\cos x+\sqrt{3} \sin x} equals
View answer and explanation
Correct option: C
12log⁡tan⁡(x2+π12)+C\frac{1}{2} \log \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)+\mathrm{C}
Option C (12log⁡tan⁡(x2+π12)+C\frac{1}{2} \log \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{12}\right)+\mathrm{C}) is correct because the integral can be simplified using trigonometric identities and the integral of cosecant.
71 Medium
71 of 411
Application +4 | -1
∫{log⁡x−11+(log⁡x)2}2dx\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2 d x is equal to
View answer and explanation
Correct option: A
x(log⁡x)2+1+C\frac{x}{(\log x)^2+1}+C
Option A (x(log⁡x)2+1+C\frac{x}{(\log x)^2+1}+C) is correct because after substitution and integration by parts, the integral simplifies to this expression.
72 Medium
72 of 411
Analyzing +4 | -1
∫(cos⁡x−cos⁡3x1−cos⁡3x)1/2dx=\int\left(\frac{\cos x-\cos ^3 x}{1-\cos ^3 x}\right)^{1 / 2} d x=
View answer and explanation
Correct option: B
−23Sin⁡−1(cos⁡3/2x)+c-\frac{2}{3} \operatorname{Sin}^{-1}\left(\cos ^{3 / 2} x\right)+c
Option C (13sin⁡3x−sin⁡x\frac{1}{3} \sin ^3 x-\sin x) is correct because after integrating and comparing with the given form, g(x) matches this expression.
73 Medium
73 of 411
Application +4 | -1
∫x3x2+2dx=\int \frac{x^3}{\sqrt{x^2+2}} d x=
View answer and explanation
Correct option: B
13(x2+2)3/2−2(x2+2)1/2+c\frac{1}{3}\left(x^2+2\right)^{3 / 2}-2\left(x^2+2\right)^{1 / 2}+c
Option B is correct because the integral is solved using a substitution, resulting in 13(x2+2)3/2−2(x2+2)1/2+c\frac{1}{3}(x^2+2)^{3/2} - 2(x^2+2)^{1/2} + c.
74 Medium
74 of 411
Application +4 | -1
The value of the integral ∫03αcosec(x−α)cosec(x−2α)dx,\int\limits_0^{3\alpha } {{\text{cosec}}\left( {x - \alpha } \right){\text{cosec}}\left( {x - 2\alpha } \right)dx} , is
View answer and explanation
Correct option: D
2    cosec    α  log⁡  (12  sec⁡  α)2\;\,\,\mathrm{cosec}\,\,\;\alpha \,\,\log \,\,\left( {\frac{1}{2}\,\,\sec \,\,\alpha } \right)
The correct answer is D (2    cosec   α  log⁡  (12  sec⁡  α)2\;\,\,\mathrm{cosec}\,\;\alpha \,\,\log \,\,\left( {\frac{{1}}{{2}}\,\,\sec \,\,\alpha } \right)) because it is the result of evaluating the definite integral using trigonometric identities and logarithmic properties.
75 Medium
75 of 411
Application +4 | -1
The value of the integral ∫0∞11  +  x4dx\int \limits_0^\infty \frac{1}{{1\,\, + \,\,{x^4}}}dx, is
View answer and explanation
Correct option: C
π2 2\frac{\pi }{{2\,\sqrt 2 }}
Option C (π22\frac{\pi}{{2\sqrt 2 }}) is correct because the integral evaluates to this value after using trigonometric substitution and simplification.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
76 Medium
76 of 411
Application +4 | -1
If ∫log⁡(x+1+x2)1+x2 dx=gof(x)+\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=g o f(x)+ Const. then
View answer and explanation
Correct option: C
f(x)=log⁡(x+x2+1)f(x)=\log \left(x+\sqrt{x^2+1}\right) and g(x)=x2/2g(x)=x^2 / 2
The correct answer is C (f(x)=log⁡(x+x2+1)f(x)=\log (x+\sqrt{x^2+1}) and g(x)=x2/2g(x)=x^2 / 2) because after substituting t=log⁡(x+1+x2)t = \log(x + \sqrt{1+x^2}), the integral becomes ∫tdt=t2/2+C\int t dt = t^2/2 + C, which matches the given form.
77 Medium
77 of 411
Application +4 | -1
Let f(x)=∫2x(x2+1)(x2+3)dxf(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x. If f(3)=12(log⁡e5−log⁡e6)f(3)=\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right), then f(4)f(4) is equal to
View answer and explanation
Correct option: C
12(log⁡e17−log⁡e19)\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right)
Option C (12(log⁡e17−log⁡e19)\frac{{1}}{{2}}(\log _{{e}} 17-\log _{{e}} 19)) is correct because after evaluating the integral and using the given condition f(3), we can find the value of f(4).
78 Medium
78 of 411
Analyzing +4 | -1
If ∫dx(x2−2x+10)2=A(tan⁡−1(x−13)+f(x)x2−2x+10)+C\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C where C\mathrm{C} is a constant of integration then :
View answer and explanation
Correct option: B
A=154A=\frac{1}{54} and f(x)=3(x−1)f(x)=3(x-1)
Option B (A=154A=\frac{{1}}{{54}} and f(x)=3(x−1)f(x)=3(x-1)) is correct because after solving the integral using trigonometric substitution, the result matches the given form with the specified values for A and f(x).
79 Medium
79 of 411
Application +4 | -1
∫cos⁡(log⁡ex)dx\int \cos \left(\log _e x\right) d x is equal to
View answer and explanation
Correct option: A
12x{cos⁡(log⁡ex)+sin⁡(log⁡ex)}\frac{1}{2} x\left\{\cos \left(\log _e x\right)+\sin \left(\log _e x\right)\right\}
Option A is correct because the integral of cos⁡(log⁡ex)\cos(\log_e x) is found using integration by parts twice, resulting in the given expression.
80 Medium
80 of 411
Application +4 | -1
The value of ∫0π∣sin⁡3  θ∣  dθ\int \nolimits_0^\pi |{\sin ^3}\,\,\theta |\,\,d\theta is
View answer and explanation
Correct option: C
43\frac{4}{3}
Option C is correct because the integral is evaluated by splitting the integral and using trigonometric substitution.
81 Medium
81 of 411
Application +4 | -1
∫tan⁡−12x1−x2dx=\int \tan ^{-1} \frac{2 x}{1-x^2} d x=
View answer and explanation
Correct option: D
2xtan⁡−1x−log⁡(1+x2)+c2 x \tan ^{-1} x-\log \left(1+x^2\right)+c
Option D (2xtan⁡−1x−log⁡(1+x2)+c2 x \tan ^{-1} x-\log \left(1+x^2\right)+c) is correct because substituting x=tan⁡θx = \tan \theta simplifies the integral, and integration by parts yields the correct result.
82 Medium
82 of 411
Analyzing +4 | -1
∫x2+n(n−1)(xsin⁡x+ncos⁡x)2dx\int \frac{x^2+n(n-1)}{(x \sin x+n \cos x)^2} d x is equal to
View answer and explanation
Correct option: D
−xnsec⁡xxnsin⁡x+nxn−1cos⁡x+tan⁡x+c\frac{-x^n \sec x}{x^n \sin x+n x^{n-1} \cos x}+\tan x+c
Option B (−sec⁡xxsin⁡x+ncos⁡x+tan⁡x+c\frac{{-\sec x}}{{x \sin x+n \cos x}}+\tan x+c) is incorrect. Option D (−xnsec⁡xxnsin⁡x+nxn−1cos⁡x+tan⁡x+c\frac{{-x^n \sec x}}{{x^n \sin x+n x^{n-1} \cos x}}+\tan x+c) is the correct answer because after integration by parts and simplification, the expression matches this form.
83 Medium
83 of 411
Application +4 | -1
If ∫1a2sin⁡2x+b2cos⁡2xdx=112tan⁡−1(3tan⁡x)+\int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+ constant, then the maximum value of asin⁡x+bcos⁡xa \sin x+b \cos x, is :
View answer and explanation
Correct option: A
40\sqrt{40}
Option A (40\sqrt{40}) is correct because after solving the integral and comparing coefficients, we find a=6 and b=2, leading to the maximum value of asin⁡x+bcos⁡xa \sin x + b \cos x as 40\sqrt{40}.
84 Medium
84 of 411
Analyzing +4 | -1
The value of the integral ∫1(x−3)6/7(x+4)8/7dx\int \frac{1}{(x-3)^{6 / 7}(x+4)^{8 / 7}} d x is
View answer and explanation
Correct option: A
(x−3x+4)1/7+C\left(\frac{x-3}{x+4}\right)^{1 / 7}+C
Option A is correct because after rewriting the integral and using a simple substitution, the integral evaluates to (x−3x+4)1/7+C(\frac{x-3}{x+4})^{1 / 7}+C.
85 Medium
85 of 411
Application +4 | -1
If f′(x)=tan⁡−1(sec⁡x+tan⁡x),x∈(−π2,π2)f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x), x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and f(0)=0f(0)=0, then the value of f(1)f(1) is
View answer and explanation
Correct option: A
π+14\frac{\pi+1}{4}
Option A is correct because the derivative simplifies to f′(x)=π4+x2f^{\prime}(x)=\frac{\pi}{4}+\frac{x}{2}, and integrating with the given condition f(0)=0f(0)=0 yields f(1)=π+14f(1)=\frac{\pi+1}{4}.
86 Medium
86 of 411
Application +4 | -1
The value of ∫log⁡x(x+1)2dx\int \frac{\log x}{(x+1)^2} d x is
View answer and explanation
Correct option: A
−log⁡xx+1+log⁡x−log⁡(x+1)\frac{-\log x}{x+1}+\log x-\log (x+1)−log⁡xx+1+log⁡x−log⁡(x+1)\frac{-\log x}{x+1}+\log x-\log (x+1)
Option A is correct because the integral of log(x)/(x+1)2 dx is found using integration by parts, resulting in -log(x)/(x+1) + log(x) - log(x+1) + c.
87 Medium
87 of 411
Application +4 | -1
The value of ∫x−ab−xdx\int \sqrt{\frac{x-a}{b-x}} d x is equal to
View answer and explanation
Correct option: B
(a−b)[sin⁡2θ2+θ]+c(a-b)\left[\frac{\sin 2 \theta}{2}+\theta\right]+c, where asin⁡2θ+bcos⁡2θ=xa \sin ^2 \theta+b \cos ^2 \theta=x
The correct answer is A because the substitution x=asin⁡2θ+bcos⁡2θx = a\sin^2 \theta + b\cos^2 \theta simplifies the integral to a form that can be easily integrated, resulting in (a−b)(sin⁡2θ2+θ)+c(a-b)(\frac{\sin 2\theta}{2} + \theta) + c.
88 Medium
88 of 411
Application +4 | -1
If ∫f(x)dx=f(x)\int f(x) d x=f(x), then ∫{f(x)}2dx\int\{f(x)\}^2 d x is equal to
View answer and explanation
Correct option: A
12{f(x)}2\frac{1}{2}\{f(x)\}^2
Option A (12{f(x)}2\frac{1}{2}\{f(x)\} ^2) is correct because by differentiating the given integral equation and solving the differential equation, we find f(x) and then integrate \{f(x)\} ^2.
89 Medium
89 of 411
Application +4 | -1
∫  1  +  tan⁡2  x1  −  tan⁡2  x    dx\int \,\,\frac{{1\,\, + \,\,{{\tan }^2}\,\,x}}{{1\,\, - \,\,{{\tan }^2}\,\,x}}\,\,\;dx is equal to
View answer and explanation
Correct option: D
12  log⁡  (1  +  tan⁡  x1  −  tan⁡  x)  +  c\frac{1}{2}\,\,\log \,\,\left( {\frac{{1\,\, + \,\,\tan \,\,x}}{{1\,\, - \,\,\tan \,\,x}}} \right)\,\, + \,\,c
The correct answer is (1/2)log((1 + tan x)/(1 - tan x)) + c because the integral simplifies to a standard form after trigonometric manipulation and substitution.
90 Medium
90 of 411
Analyzing +4 | -1
If ∫sin⁡  θ  −  cos⁡  θ(sin⁡  θ  +  cos⁡  θ)  sin⁡  θ  cos⁡  θ  +  sin⁡2  θ  cos⁡2  θdθ=cosec−1  (f(θ))+C\int \frac{{\sin \,\,\theta \,\, - \,\,\cos \,\,\theta }}{{\left( {\sin \,\,\theta \,\, + \,\,\cos \,\,\theta } \right)\,\,\sqrt {\sin \,\,\theta \,\,\cos \,\,\theta \,\, + \,\,{{\sin }^2}\,\,\theta \,\,{{\cos }^2}\,\,\theta } }}d\theta = \mathrm{cose}{\mathrm{c}^{ - 1}}\,\,\left( {f\left( \theta \right)} \right) + C, then
View answer and explanation
Correct option: A
f(θ)=sin⁡  2θ  +  1f\left( \theta \right) = \sin \,\,2\theta \,\, + \,\,1
Option A (f(θ)=sin⁡ 2θ + 1f(\theta ) = \sin \,2\theta \, + \,1) is correct because after simplifying and integrating, the result matches the given form with this f(θ)f(\theta).
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