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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Easy

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301 Easy
301 of 746
Application +4 | -1
∫1sin⁡(x−a)sin⁡(x−b)dx=\int \frac{1}{\sin (x-a) \sin (x-b)} d x=
View answer and explanation
Correct option: A
1sin⁡(a−b)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C
Option A (1sin⁡(a−b)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C) is correct because by using the sine subtraction formula and simplifying, the integral evaluates to this expression.
302 Easy
302 of 746
Application +4 | -1
The area bounded by the curve y=sin⁡xy=\sin x from 0 to π\pi and x\mathrm{x}-axis is sq. units.
View answer and explanation
Correct option: A
2
Option A (9) is correct because the area under the curve y = x2 from x = 0 to x = 3 is calculated by integrating x2 from 0 to 3.
303 Easy
303 of 746
Application +4 | -1
If f(x)=cos⁡xf(x)=\cos x then ∫(2(f(x))2−1)(4(f(x))3−3\int\left(2(f(x))^2-1\right)\left(4(f(x))^3-3\right. f(x))dxf(x)) \mathrm{d} x is equal to
View answer and explanation
Correct option: A
110sin⁡5x+12sin⁡x+C\frac{1}{10} \sin 5 x+\frac{1}{2} \sin x+C
The correct answer is A because using trigonometric identities to simplify the integrand and then integrating yields the result.
304 Easy
304 of 746
Analyzing +4 | -1
Let f:R→Rf:R \to R be a differentiable function such that f(π4)=2,  f(π2)=0f\left( {\frac{\pi }{4}} \right) = \sqrt 2 ,\,\,f\left( {\frac{\pi }{2}} \right) = 0 and f′(π2)=1f'\left( {\frac{\pi }{2}} \right) = 1 and let g(x)=∫xπ4(f′(t)sec⁡t+tan⁡tsec⁡t(t))dtg\left( x \right) = \int_x^{\frac{\pi }{4}} {\left( {f'\left( t \right)\sec t + \tan t\sec t\left( t \right)} \right)} dt for x∈[π4,π2).x \in \left[ {\frac{\pi }{4},\frac{\pi }{2}} \right). Then lim⁡x→(π2)−g(x)\lim\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ - }} g\left( x \right) is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because the limit of g(x)g(x) as xx approaches π2\frac{\pi}{2} from the left is found by substituting x=π2−hx = \frac{\pi}{2} - h and applying L'Hôpital's rule, resulting in 3.
305 Easy
305 of 746
Application +4 | -1
∫sin⁡−1(3x−4x3)dx=\int \sin ^{-1}\left(3 x-4 x^3\right) d x=
View answer and explanation
Correct option: D
3[xsin⁡−1x+1−x2]+c3\left[x \sin ^{-1} x+\sqrt{1-x^2}\right]+c
Option D (3[xsin⁡−1x+1−x2]+c3\left[x \sin ^{-1} x+\sqrt{1-x^2}\right]+c) is correct because substituting x=sin⁡θx = \sin \theta simplifies the integral, and integration by parts yields the correct result.
306 Easy
306 of 746
Application +4 | -1
The value of ∫sin⁡nxcos⁡n+6x3dx\int \sqrt[3]{\frac{\sin ^n x}{\cos ^{n+6} x}} d x, is
View answer and explanation
Correct option: B
3n+3tan⁡n/3+1x+C\frac{3}{n+3} \tan ^{n / 3+1} x+C
Option B is correct because the integral simplifies to (3/(n+3))tan(n/3 + 1)(x) + C after recognizing the derivative of tan(x).
307 Easy
307 of 746
Application +4 | -1
∫[sin⁡(log⁡x)+cos⁡(log⁡x)]dx=\int[\sin (\log x)+\cos (\log x)] d x=
View answer and explanation
Correct option: D
xsin⁡(log⁡x)+cx \sin (\log x)+c
Option D (xsin⁡(log⁡x)+cx \sin (\log x)+c) is correct because it is the result of integrating the given expression using integration by parts.
308 Easy
308 of 746
Application +4 | -1
If ∫(sin⁡2x+cos⁡2x)dx=12sin⁡(2x−c)+a\int(\sin 2 x+\cos 2 x) d x=\frac{1}{\sqrt{2}} \sin (2 x-c)+a, then the value of aa and cc is
View answer and explanation
Correct option: A
c=π/4c=\pi / 4 and a=ka=k (an arbitrary constant)
Option A is correct because the integral of sin(2x) + cos(2x) can be expressed in the form (1/sqrt(2))sin(2x - c) + a, where c = pi/4 and a is an arbitrary constant.
309 Easy
309 of 746
Application +4 | -1
∫1x1−x3dx\int \frac{1}{x \sqrt{1-x^3}} d x is equal to
View answer and explanation
Correct option: A
13log⁡∣1−x3−11−x3+1∣+C\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C
Option A (13log⁡∣1−x3−11−x3+1∣+C\frac{{1}}{{3}} \log \left| \frac{{\sqrt{{1-x^3}}-1}}{{\sqrt{{1-x^3}}+1}} \right|+C) is correct because it is the result of the integration after appropriate substitution and simplification.
310 Easy
310 of 746
Application +4 | -1
Let f  (x)=∫  1(1+x2)3/2  dxf\,\,\left( x \right) = \int \,\,\frac{1}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,dx and f  (0)=0f\,\,\left( 0 \right) = 0, then f  (1)=f\,\,\left( 1 \right) =
View answer and explanation
Correct option: B
12\frac{1}{{\sqrt 2 }}
Option B (12\frac{1}{{\sqrt 2 }}) is correct because the integral of the given function is evaluated, and the constant of integration is determined using the initial condition f(0) = 0.
311 Easy
311 of 746
Application +4 | -1
∫x2(3)x3+1dx=\int x^2(3)^{x^3+1} d x=
View answer and explanation
Correct option: B
(3)x3log⁡3+c\frac{(3)^{x^3}}{\log 3}+c
Option B is correct because the integral can be solved using u-substitution and the integral of an exponential function, resulting in (3)x3log⁡3+c\frac{(3)^{x^3}}{\log 3}+c.
312 Easy
312 of 746
Analyzing +4 | -1
The integral ∫1(1+x)x−x2dx\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x is equal to (where CC is the constant of integration)
View answer and explanation
Correct option: B
−21−x1+x-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Option B is correct because after using the trigonometric substitution x=cos⁡2θx = \cos^2 \theta, the integral simplifies and evaluates to −21−x1+x+C-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C.
313 Easy
313 of 746
Analyzing +4 | -1
∫  x2+  1  [log⁡  (x2+1)−2  log⁡  x]x4  dx\int \,\,\frac{{\sqrt {{x^2} + \,\,1} \,\,\left[ {\log \,\,\left( {{x^2} + 1} \right) - 2\,\,\log \,\,x} \right]}}{{{x^4}}}\,\,dx is equal to
View answer and explanation
Correct option: B
−13  (1  +  1x2)3/2  [log⁡  (1  +  1x2)  −  23]+C- \frac{1}{3}\,\,{\left( {1\,\, + \,\,\frac{1}{{{x^2}}}} \right)^{3/2}}\,\,\left[ {\log \,\,\left( {1\,\, + \,\,\frac{1}{{{x^2}}}} \right)\,\, - \,\,\frac{2}{3}} \right] + C
Option B (−13 (1 + 1x2)3/2  [log⁡  (1 + 1x2) −  23]+C- \frac{1}{3}\,{\left( {1\, + \,\frac{1}{{{{x^2}}}}} \right)^{3/2}}\,\,\left[ {\log \,\,\left( {1\, + \,\frac{1}{{{{x^2}}}}} \right)\, - \,\,\frac{2}{3}} \right] + C) is correct because after substitution and integration by parts, the integral evaluates to the given expression.
314 Easy
314 of 746
Application +4 | -1
∫xlog⁡xdx=\int x \log x d x=
View answer and explanation
Correct option: B
x22log⁡x−x24+c\frac{x^2}{2} \log x-\frac{x^2}{4}+c
Option B (x22log⁡x−x24+c\frac{x^2}{2} \log x-\frac{x^2}{4}+c) is correct because it is the result of integrating xlog⁡xx \log x using integration by parts.
315 Easy
315 of 746
Analyzing +4 | -1
If ∫sin⁡−1(x1+x)dx = A(x) tan⁡−1(x) + B (x) + C, where C\int {{{\sin }^{ - 1}}\left( {\sqrt {\frac{x}{{1 + x}}} } \right)dx\, = \,A(x)\,{{\tan }^{ - 1}}\left( {\sqrt x } \right)\, + \,B\,(x)\, + \,C,\,\mathrm{where}\,C} is a constant of integration, then the ordered pair (A(x), B (x))\left( {A(x),\,B\,(x)} \right) can be
View answer and explanation
Correct option: C
(x+1, −x)\left( {x + 1,\, - \sqrt x } \right)
Option C ((x+1, −x)\left( {x + 1,\, - \sqrt x } \right)) is correct because after evaluating the integral using trigonometric substitution and integration by parts, the result matches the given form with A(x) = x+1 and B(x) = -x\sqrt{x}.

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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
316 Easy
316 of 746
Application +4 | -1
∫−π/4π/4dx1+cos⁡2x\int\limits_{{-\pi / 4}}^{\pi / 4} \frac{d x}{1+\cos 2 x} is equal to
View answer and explanation
Correct option: A
1
Option A (1) is correct because the integral simplifies to the integral of sec²(x) from 0 to π/4, which equals tan(π/4) - tan(0) = 1.
317 Easy
317 of 746
Application +4 | -1
If ∫(1−tan⁡3x)2dx=13[tan⁡3x+log⁡f(x)]+C\int(1-\tan 3 x)^2 d x=\frac{1}{3}[\tan 3 x+\log f(x)]+C then f(x)f(x) is given by
View answer and explanation
Correct option: B
cos⁡23x\cos ^2 3 x
Option B (cos⁡23x\cos ^2 3 x) is correct because after expanding and integrating, the result matches the given form with f(x)=cos⁡23xf(x) = \cos^2 3x.
318 Easy
318 of 746
Application +4 | -1
If the antiderivative of 1x+x3/2\frac{1}{\sqrt{x+x^{3 / 2}}} is A1+x+CA \sqrt{1+\sqrt{x}}+C then AA is equal to
View answer and explanation
Correct option: D
4
Option D (4) is correct. By substituting \x1/2=tx^{1/2} = t, the integral simplifies, leading to the antiderivative \41+x+C4 \sqrt{1+\sqrt{x}} + C, thus A = 4.
319 Easy
319 of 746
Analyzing +4 | -1
If ∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2xdx=asin⁡2x+C\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=a \sin 2 x+C, then a=a=
View answer and explanation
Correct option: A
−12-\frac{1}{2}
Option A is correct because after simplifying the integral, we get −12sin⁡2x+C-\frac{1}{2} \sin 2x + C, which means a=−12a = -\frac{1}{2}.
320 Easy
320 of 746
Application +4 | -1
∫tan⁡−1(1−x1+x)dx\int \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right) d x can be evaluated by the subsititution
View answer and explanation
Correct option: B
x=cos⁡θx=\cos \theta
Option B (x=cos⁡θx=\cos \theta) is correct because the substitution x=cos⁡θx = \cos \theta simplifies the expression inside the inverse tangent to tan⁡θ2\tan \frac{\theta}{2}, making the integral easier to evaluate.
321 Easy
321 of 746
Application +4 | -1
∫1sin⁡x−cos⁡x+2dx\int \frac{1}{\sin x-\cos x+\sqrt{2}} d x equals
View answer and explanation
Correct option: D
−12cot⁡(x2+π8)-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)
Option D (−12cot⁡(x2+π8)-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)) is correct because it is the result of integrating the given expression after trigonometric simplification.
322 Easy
322 of 746
Analyzing +4 | -1
∫  x2−1x4+x2+1  dx\int \,\,\frac{{{x^2} - 1}}{{{x^4} + {x^2} + 1}}\,\,dx is equal to
View answer and explanation
Correct option: C
12  log⁡  x2  −  x  +  1x2  +  x  +  1  +  c\frac{1}{2}\,\,\log \,\,\frac{{{x^2}\,\, - \,\,x\,\, + \,\,1}}{{{x^2}\,\, + \,\,x\,\, + \,\,1}}\,\, + \,\,c
Option C (12  log⁡  x2 − x + 1x2 + x + 1 + c\frac{1}{2}\,\,\log \,\,\frac{{{x^2}\, - \,x\, + \,1}}{{{x^2}\, + \,x\, + \,1}}\, + \,c) is correct because after dividing by x2x^2 and using substitution, the integral can be solved to obtain this result.
323 Easy
323 of 746
Application +4 | -1
The value of ∫x2−1(x2+1)x4+1dx\int \frac{x^2-1}{\left(x^2+1\right) \sqrt{x^4+1}} d x, is equal to
View answer and explanation
Correct option: A
12sec⁡−1(x2+12x)+C\frac{1}{\sqrt{2}} \sec ^{-1}\left(\frac{x^2+1}{\sqrt{2} x}\right)+C
Option A (12sec⁡−1(x2+12x)+C\frac{{1}}{{\sqrt{{2}}}} \sec ^{-1}(\frac{{x^2+1}}{{\sqrt{{2}} x}})+C) is correct because the integral simplifies to this expression after dividing by x2x^2 and using a suitable substitution.
324 Easy
324 of 746
Application +4 | -1
∫  cos⁡  x−  sin⁡  x1+2  sin⁡  x  cos⁡  x  dx\int \,\,\frac{{\cos \,\,x - \,\,\sin \,\,x}}{{1 + 2\,\,\sin \,\,x\,\,\cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: C
−  1sin⁡  x  +  cos⁡  x  +  c- \,\,\frac{1}{{\sin \,\,x\,\, + \,\,\cos \,\,x}}\,\, + \,\,c
Option C ( −  1sin⁡ x +  cos⁡ x + c\ - \,\,\frac{1}{{\sin \,x\, + \,\,\cos \,x}}\, + \,c) is correct because it is the result of a u-substitution where u=sin⁡x+cos⁡xu = \sin x + \cos x.
325 Easy
325 of 746
Application +4 | -1
∫x2−1x32x4−2x2+1dx\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x is equal to
View answer and explanation
Correct option: D
2x4−2x2+12x2+C\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C
Option D is correct because after substitution and simplification, the integral can be solved, resulting in 2x4−2x2+12x2+C\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C.
326 Easy
326 of 746
Application +4 | -1
The integral ∫(1+x−1x)ex+1xdx\int\left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} d x is equal to
View answer and explanation
Correct option: D
xex+1x+Cx e^{x+\frac{1}{x}}+C
Option D (xex+1x+Cx e^{x+\frac{1}{x}}+C) is correct because integrating by parts simplifies the expression to the given result.
327 Easy
327 of 746
Application +4 | -1
∫xsec⁡2xdx=\int x \sec ^2 x d x=
View answer and explanation
Correct option: D
xtan⁡x+log⁡cos⁡x+cx \tan x+\log \cos x+c
Option D (xtan⁡x+log⁡cos⁡x+cx \tan x+\log \cos x+c) is correct because it is the result of integrating xsec⁡2xx \sec ^2 x using integration by parts.
328 Easy
328 of 746
Application +4 | -1
If ∫f(x)dx=2{f(x)}3+C\int f(x) d x=2\{f(x)\}^3+C, then f(x)f(x) is
View answer and explanation
Correct option: D
x3\sqrt{\frac{x}{3}}
Option D is correct. Differentiating both sides of the given equation and solving the resulting differential equation yields the solution.
329 Easy
329 of 746
Application +4 | -1
∫cos⁡−3/7x  sin⁡−11/7x dx\int {\cos ^{ - 3/7}}x\,\,{\sin ^{ - 11/7}}x\,dx is equal to
View answer and explanation
Correct option: C
−  74  tan⁡−4/7x+c- \,\,\frac{7}{4}\,\,{\tan ^{ - 4/7}}x + c
Option C (− 74 tan⁡−4/7x+c- \,\frac{7}{4}\,{{\tan }^{ - 4/7}}x + c) is correct because it is the result of integrating the given expression after simplification and substitution.
330 Easy
330 of 746
Application +4 | -1
∫(ex+e−x)2⋅(ex−e−x)dx\int\left(e^x+e^{-x}\right)^2 \cdot\left(e^x-e^{-x}\right) d x is equal to
View answer and explanation
Correct option: D
13(ex+e−x)3+C\frac{1}{3}\left(e^x+e^{-x}\right)^3+C
Option D (13(ex+e−x)3+C\frac{{1}}{{3}}\left(e^x+e^{-x}\right)^3+C) is correct because by substituting t=ex+e−xt = e^x + e^{-x}, the integral simplifies to 13t3+C\frac{{1}}{{3}}t^3 + C.
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