The value of $\mathop \smallint \limits_0^{\pi /2} \,\,\frac{{\sin \,\,8x\,\,\log \,\,\cot \,\,x}}{{\cos \,\,2x}}\,\,dx$, is

$\frac{{{\rm{5}}\,{\rm{\pi }}}}{{\rm{2}}}$

$\frac{{{\rm{3}}\,{\rm{\pi }}}}{{\rm{2}}}$

If $I\left( {m,\;\,n} \right) = \mathop \smallint \limits_0^1 \,{x^{m - 1}}{\left( {1 - x} \right)^{n - 1}}\,dx$, then

$I\left( {m,\;\,n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^{m - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^{n - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx$

$I\left( {m,\,\;n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^m}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^n}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx$

$I\left( {m,\;n} \right) = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^n}}}{{{{\left( {1 + x} \right)}^{m + n - 1}}}}\,\,dx = \mathop \smallint \limits_0^\infty \,\,\frac{{{x^m}}}{{{{\left( {1 + x} \right)}^{m + n - 1}}}}\,\,dx$

If $F\left( x \right) = \mathop \smallint \nolimits_{{x^2}}^{{x^3}} \,\,\log \,\,t\,dt,\;\,\left( {x > 0} \right)$, then $F'\,\left( x \right)$ is equal to

$\left( {4x\,\, - \,\,9{x^2}} \right)\log \,\,x$

$\frac{d}{{dx}}\,\,\mathop \smallint \nolimits_{f\left( x \right)}^{{\rm{g}}\left( x \right)} \,\,{\rm{h}}\,\,\left( t \right)\,\,dt$ is equal to

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]$

$g'\left( x \right)\,\,h\,\,\left[ {{\rm{g}}\,\,\left( x \right)} \right]$

$h\,\,\left[ {\,{\rm{g}}\,\,\left( x \right)} \right]\,\, - \,\,h\left[ {\,f\,\,\left( x \right)} \right]$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 91-120 of 791 questions

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91 Easy

91 of 791

Application +4 | -1

\int_0^{20 \pi}(|\sin x|+|\cos x|)^2 d x

is equal to

View answer and explanation

Correct option: D

20(\pi+2)

The correct answer is D (

20(\pi+2)

) because it is the result of evaluating the definite integral

\int_0^{20 \pi}(|\sin x|+|\cos x|)^2 dx

as shown in the step-by-step solution.

92 Easy

92 of 791

Application +4 | -1

\int \nolimits_0^\pi \,\,\frac{{x\,\,dx}}{{1\,\, + \,\,\sin \,\,x}}\,\,\mathrm{is }\,\mathrm{equal}\,\mathrm{ to}

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because after applying the property of definite integrals and simplifying, the integral evaluates to

\pi

93 Easy

93 of 791

Application +4 | -1

The value of

\int\limits_{{1 / \pi}}^{2 / \pi} \frac{\sin 1 / x}{x^2} \mathrm{~d} x

is equal

View answer and explanation

Correct option: A

Option A (1) is correct because substituting

t = \frac{{1}}{{x}}

transforms the integral to

\int_{\pi/2}^{\pi} \sin(t) dt

, which evaluates to 1.

94 Easy

94 of 791

Application +4 | -1

\int\limits_{0}^1 \frac{\sqrt{e^x}}{\sqrt{e^x+e^{-x}}} \mathrm{~d} x=\log \frac{e+A}{1+\sqrt{2}}

then

A

is equal to

View answer and explanation

Correct option: B

\sqrt{e^2+1}

Option B (

\sqrt{{e^2+1}}

) is correct because after substituting

e^x = t

, the integral evaluates to

\log \frac{{e+\sqrt{{e^2+1}}}}{{1+\sqrt{{2}}}}

, thus

A = \sqrt{{e^2+1}}

95 Easy

95 of 791

Analyzing +4 | -1

\int\limits_{ - \pi }^\pi {|\pi - |x||dx}

is equal to

View answer and explanation

Correct option: C

{\pi ^2}

Option C

{\pi ^2}

is correct because the definite integral evaluates to this value after considering the absolute value function.

96 Easy

96 of 791

Application +4 | -1

The area bounded by

y = - {x^2} + 2x + 3\;\mathrm{and}\;y = 0\;\mathrm{is}

View answer and explanation

Correct option: B

\frac{{32}}{3}

sq units

Option B (32/3 sq units) is correct because it is the result of integrating the quadratic function between its roots.

97 Easy

97 of 791

Analyzing +4 | -1

\int_{-1 / \sqrt{3}}^{1 / \sqrt{3}} \frac{x^4}{1-x^4} \cos ^{-1} \frac{2 x}{1+x^2} d x=k \int_0^{1 / \sqrt{3}} \frac{x^4}{1-x^4}

, then '

k

' is equal to

View answer and explanation

Correct option: A

\pi

Option A (

\pi

) is correct because, after applying King's property and simplifying, the value of k is found to be

\pi

98 Easy

98 of 791

Application +4 | -1

For

k \geq 0, \lim \limits_{{n \rightarrow \infty}} \frac{1^k+3^k+\ldots .+(2 n-1)^k}{n^{k+1}}=

View answer and explanation

Correct option: B

\frac{2^k}{k+1}

Option B (

\frac{2^k}{k+1}

) is correct because the limit of the sum can be expressed as a definite integral, which evaluates to

\frac{2^k}{k+1}

99 Easy

99 of 791

Application +4 | -1

A polynomial

P

is positive for

x > 0

, and the area of the region bounded by

P(x)

, the

x

-axis, and the vertical lines

x=0

and

x=K

K^2(K+3) / 3

. The polynomial

P(x)

View answer and explanation

Correct option: A

x^2+2 x

Option A (

x^2+2x

) is correct because differentiating the given area function with respect to K yields the polynomial P(K), which simplifies to

x^2+2x

100 Easy

100 of 791

Application +4 | -1

\int \nolimits_0^{\pi /2} \,\,\frac{{\sqrt {\cot \,\,x} }}{{\sqrt {\cot \,\,x\,\,} + \,\,\sqrt {\tan \,\,x} }}\,\,dx

is equal to

View answer and explanation

Correct option: D

\frac{\mathrm{\pi }}{4}

Option D (

\frac{\mathrm{\pi }}{4}

) is correct because the integral can be solved using the property of definite integrals where

\int \nolimits_0^a f(x) dx = \int \nolimits_0^a f(a-x) dx

101 Easy

101 of 791

Analyzing +4 | -1

The value of

\int \nolimits_0^{\pi /2} \,\,\frac{{{{\sin }^2}\,\,x\,\, - \,\,{{\cos }^2}\,\,x}}{{{{\sin }^3}\,\,x\,\, + \,\,{{\cos }^3}\,\,x}}\,\,dx

View answer and explanation

Correct option: A

\mathrm{0}

Option A is correct because the integral evaluates to 0 due to the symmetry and properties of definite integrals.

102 Easy

102 of 791

Application +4 | -1

\int \nolimits_{ - \pi /2}^{\pi /2} \,\,\frac{{dx}}{{1 + \cos \,\,x}}\,\,\mathrm{is}

is equal to

View answer and explanation

Correct option: C

\mathrm{2}

Option C (2) is correct because the integral simplifies to

\int \nolimits_0^{\pi /2} {\sec ^2}\,\frac{x}{2}\,dx

which evaluates to 2.

103 Easy

103 of 791

Application +4 | -1

The value of

I = \int \nolimits_0^1 \,\,x\,\,\left| {\,x\,\, - \,\,\frac{1}{2}\,} \right|\,\,dx

View answer and explanation

Correct option: C

\mathrm{1}\,\mathrm{/}\,\mathrm{8}

Option C (

\mathrm{1}\,\mathrm{/}\,\mathrm{8}

) is correct because the definite integral is evaluated by splitting the integral at x=1/2 due to the absolute value function.

104 Easy

104 of 791

Application +4 | -1

Evaluate

\int_0^3\left(\mathrm{a}^2-\mathrm{x}^2\right)^{5 / 2} \mathrm{dx}

View answer and explanation

Correct option: A

\frac{5 \pi a^6}{32}

Option A (

\frac{{5 \pi a^6}}{{32}}

) is correct because the integral is solved using trigonometric substitution and reduction formulas.

105 Easy

105 of 791

Application +4 | -1

f(x)

and

g(x)

are two continuous functions defined on

[-a, a]

, then the value of

\int_{ - a}^a {\{ f(} x) + f( - x)\} \{ g(x) - g( - x)\} \,\,dx\mathrm{, }\,\,\mathrm{is }

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the integrand is an odd function, making the integral over a symmetric interval equal to zero.

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SEQUENCE OF SERIES 1764 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

106 Easy

106 of 791

Analyzing +4 | -1

The value of

\int \limits_0^{\pi /2} \,\,\frac{{\sin \,\,8x\,\,\log \,\,\cot \,\,x}}{{\cos \,\,2x}}\,\,dx

, is

View answer and explanation

Correct option: A

0

Option A (0) is correct because the integrand is an odd function, and the integral of an odd function over a symmetric interval is zero.

107 Easy

107 of 791

Application +4 | -1

The ratio of the area bounded by the curves

y=\cos x

and

y=\cos 2 x

between

x=0, x=\pi / 3

and

x

-axis, is

View answer and explanation

Correct option: D

2: 1

Option D (2:1) is correct because the ratio of the areas is calculated by integrating the given functions over the specified interval and then dividing the results.

108 Easy

108 of 791

Application +4 | -1

\int\limits_{0}^\pi \sqrt{\frac{1+\cos 2 x}{2}} d x=

View answer and explanation

Correct option: A

Option A (2) is correct because the integral simplifies to the integral of the absolute value of cos(x) from 0 to pi, which equals 2.

109 Easy

109 of 791

Application +4 | -1

\int\limits_{0}^\pi x \sin x d x=

View answer and explanation

Correct option: A

\pi

Option A (

\pi

) is correct because the integral of

x \sin x

from 0 to

\pi

is equal to

\pi

after applying integration by parts and using the properties of definite integrals.

110 Easy

110 of 791

Application +4 | -1

I\left( {m,\;\,n} \right) = \int \limits_0^1 \,{x^{m - 1}}{\left( {1 - x} \right)^{n - 1}}\,dx

, then

View answer and explanation

Correct option: A

I\left( {m,\;\,n} \right) = \int \limits_0^\infty \,\,\frac{{{x^{m - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx = \int \limits_0^\infty \,\,\frac{{{x^{n - 1}}}}{{{{\left( {1 + x} \right)}^{m + n}}}}\,\,dx

Option A is correct because it represents the Beta function in terms of an integral from 0 to infinity, derived through substitution.

111 Easy

111 of 791

Application +4 | -1

F\left( x \right) = \int \nolimits_{{x^2}}^{{x^3}} \,\,\log \,\,t\,dt,\;\,\left( {x > 0} \right)

, then

F'\,\left( x \right)

is equal to

View answer and explanation

Correct option: A

\left( {9{x^2} - \,\,4x} \right)\,\,\log \,\,x

Option A (

\left( {9{x^2} - \,4x} \right)\,\,\log \,x

) is correct because it is the result of applying Leibniz's rule for differentiation under the integral sign.

112 Easy

112 of 791

Application +4 | -1

\int\limits_{0}^{\frac{\pi}{3}} \cos ^4 x d x=a \pi+b \sqrt{3}

, where

a

and

b

are rational numbers, then

9 a+8 b

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because it is the value of the expression 9a + 8b, where a and b are the rational coefficients obtained after evaluating the definite integral of cos⁴(x) from 0 to pi/3.

113 Easy

113 of 791

Application +4 | -1

\frac{d}{{dx}}\,\,\int \nolimits_{f\left( x \right)}^{\mathrm{g}\left( x \right)} \,\,\mathrm{h}\,\,\left( t \right)\,\,dt

is equal to

View answer and explanation

Correct option: C

g'\left( x \right)\,\,h\,\,\left[ {\mathrm{g}\,\,\left( x \right)} \right]\,\, - \,\,f'\left( x \right)\,\,h\,\,\left[ {f\left( x \right)} \right]

Option C (

g'\left( x \right)\\,,h\\,\left[ {\mathrm{g}\\,,\left( x \right)} \right]\\, - \\,f'\left( x \right)\\,,h\\,\left[ {f\left( x \right)} \right]

) is correct because it represents the Leibniz rule for differentiating an integral with variable limits.

114 Easy

114 of 791

Application +4 | -1

f(x)=\left\{\begin{array}{lll}4 x+3, & \text { if } & 1 \leq x \leq 2 \\ 3 x+5, & \text { if } & 2 < x \leq 4\end{array}\right.

then

\int\limits_{1}^4 f(x) d x=

View answer and explanation

Correct option: D

Option D (37) is correct because it is the result of evaluating the piecewise defined function over the given interval by splitting the integral at the point where the function definition changes.

115 Easy

115 of 791

Application +4 | -1

The value of

\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{{{\sin }^2}x}}{{1 + {2^x}}}dx}

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D (

\frac{\pi }{4}

) is correct because the integral is solved by using the property of definite integrals and combining the integral with a modified form to simplify and evaluate.

116 Easy

116 of 791

Application +4 | -1

\int\limits_{{\pi / 4}}^{3 \pi / 4} \frac{1}{1+\cos x} d x

is equal to

View answer and explanation

Correct option: A

Option A is correct. The integral is evaluated by using trigonometric identities to simplify the integrand and then integrating.

117 Easy

117 of 791

Application +4 | -1

The value of

\int\limits_{0}^{\pi / 2} \frac{\mathrm{d} x}{1+\tan ^3 x}

View answer and explanation

Correct option: C

\pi / 4

Option C (π/4) is correct because the integral is evaluated using the property ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a-x) dx and adding the two forms of the integral.

118 Easy

118 of 791

Application +4 | -1

The value of

\int\limits_{0}^\pi|\cos x|^3 d x

is :

View answer and explanation

Correct option: D

\frac{4}{3}

Option D (

\frac{{4}}{{3}}

) is correct because the integral of |cos(x)|^3 from 0 to pi is twice the integral from 0 to pi/2, which evaluates to 4/3.

119 Easy

119 of 791

Application +4 | -1

The value of

\int\limits_{{0}}^{n^{2}}[\sqrt{x}] d x,([*]

denotes the greatest integer function),

n \in N

, is

View answer and explanation

Correct option: B

\frac{1}{6} n(n-1)(4 n+1)

Option B is correct because the integral is evaluated by splitting it into intervals where the greatest integer function is constant.

120 Easy

120 of 791

Application +4 | -1

\int\limits_{0}^{\pi / 4} \frac{x \sin x}{\cos ^3 x} \mathrm{~d} x=\frac{\pi}{4}+A

, then

A

is equal to

View answer and explanation

Correct option: D

-\frac{1}{2}

Option D (

-\frac{1}{2}

) is correct because integration by parts and trigonometric identities are used to evaluate the definite integral, resulting in the given value.

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