The value of the integral $\mathop \smallint \limits_0^\pi \,\,\frac{{x\,dx}}{{1 + \cos \,\,\alpha \,\,\sin \,\,x}},\;\,0 < \alpha < \pi $, is

$\frac{{\pi \alpha }}{{1 + \sin \,\,\alpha }}$

$\frac{{\pi \alpha }}{{\cos \,\,\alpha }}$

$\frac{{\pi \alpha }}{{1 + \cos \,\,\alpha }}$

The value of the integral $\mathop \smallint \limits_0^\pi \,\,\frac{1}{{{a^2} - \,\,2a\,\,\cos \,\,x\,\, + \,\,1}}\,\,dx\,\,(a < 1)$ is

$\frac{\pi }{{{a^2}\,\, - \,\,1}}$

$\frac{{2\;\pi }}{{{a^2}\,\, - \,\,1}}$

If $f\,\,\left( x \right) = \mathop \smallint \nolimits_{{x^2}}^{{x^2} + 1} \,\,{e^{ - {t^2}}}\,dt$, then $f\,\,\left( x \right)$ increases in

$\left( {0,\;\,\,\infty } \right)$

If $(\,\mathop \smallint \nolimits_0^a \,\,x\,dx\,)\;\,\, \le \,\,\,\left( {a + 4} \right)$, then

$a\,\,\, \le \,\, - 2$ or $a\,\,\, \ge \,\,4$

If $\smallint \,\,\frac{{{x^5}}}{{\sqrt {1 + {x^3}} }}\,\,dx$ is equal to

$\frac{2}{9}\,\,{\left( {1 + {x^3}} \right)^{5/2}}\,\, + \,\,\frac{2}{3}\,\,{\left( {1 + {x^3}} \right)^{3/2}}\,\, + \,\,C$

$\log \,\,\left| {\sqrt x + \sqrt {1 + {x^3}} } \right|\, + \,\,C$

${x^2}\,\,\log \,\,\left( {1 + {x^3}} \right)\,\, + \,\,C$

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JEE • MATHEMATICS • DEFINITE INTEGRATION

Definite integration practice · Mixed difficulty

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871 Medium

871 of 995

Application +4 | -1

\int \nolimits_{\pi /6}^{\pi /3} \,\,\frac{{dx}}{{1 + \sqrt {\tan \,\,x} }}

is equal to

View answer and explanation

Correct option: A

\frac{\pi }{{12}}

Option A (

\frac{\pi }{{12}}

) is correct because using the property of definite integrals

\int \nolimits_a^b f(x) dx = \int \nolimits_a^b f(a + b - x) dx

simplifies the integral to

\frac{\pi}{12}

872 Easy

872 of 995

Application +4 | -1

f:R \to R

and

\mathrm{g}:R \to R

are one to one, real valued functions, then the value of the integral

\int \nolimits_{ - \pi }^\pi \,\,\,\left[ {f\left( x \right)\,\, + \,\,f\,\left( { - x} \right)} \right]\,\,\left[ {\mathrm{g}\,\,\left( x \right)\,\, - \,\,\mathrm{g}\,\,\left( { - x} \right)} \right]\,\,dx

View answer and explanation

Correct option: A

\mathrm{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.

873 Easy

873 of 995

Application +4 | -1

\lim \limits_{{n \rightarrow \infty}} \frac{1}{n} \sum\limits_{{r=1}}^{2 n} \frac{r}{\sqrt{n^2+r^2}}

equals

View answer and explanation

Correct option: B

-1+\sqrt{5}

Option B (

-1+\sqrt{5}

) is correct because the limit of the Riemann sum is equivalent to a definite integral.

874 Hard

874 of 995

Analyzing +4 | -1

Let

f

be a differentiable function from

R

R

such that

\left| {f(x) - f(y)} \right| \le 2{\left| {x - y} \right|^{\frac{3}{2}}},

for all

x,y \in R.

If f

(0) = 1

, then

\mathrm{ }\int\limits_0^1 {{f^2}(x)dx}

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because the given condition implies that f'(x) = 0, making f(x) a constant function. Since f(0) = 1, f(x) = 1 for all x, and the integral of f²(x) from 0 to 1 is 1.

875 Easy

875 of 995

Analyzing +4 | -1

f\left( x \right)

is defined

\left[ { - \,2,\;\,2} \right]\mathrm{\;}\,\,\mathrm{by\;}\,\,f\left( x \right) = 4{x^2} - \,\,3x + 1

and

g\left( x \right) = \frac{{f\left( { - x} \right)\,\, - \,\,f\left( x \right)}}{{{x^2} + \,\,3}}

,
then

\int \nolimits_{ - 2}^2 \,\,\mathrm{g}\,\,\left( x \right)dx

is equal to

View answer and explanation

Correct option: C

\mathrm{0}

The correct answer is C (0) because g(x) is an odd function, and the integral of an odd function over a symmetric interval is zero.

876 Easy

876 of 995

Analyzing +4 | -1

The value of the integral

\int \limits_{ - \pi /2}^{\pi /2} \log \,\,\,\left( {\frac{{a\,\, - \,\,\sin \,\,\theta }}{{a\,\, + \,\,\sin \,\,\theta }}} \right)\,\,d\,\,\theta ,\,\,a > 0

, is

View answer and explanation

Correct option: A

\mathrm{0}

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

877 Easy

877 of 995

Application +4 | -1

\int \nolimits_1^3 \,\,\frac{{\cos \,\,(\log \,\,x)}}{x}\;\,\,dx

is equal to

View answer and explanation

Correct option: C

\sin \,(\log \,\,3)\,

Option C (

\sin \,(\log \,\,3)\,

) is correct because the integral is solved using a simple substitution, recognizing that the derivative of log(x) is 1/x.

878 Medium

878 of 995

Application +4 | -1

The value of the integral

\int \limits_0^\pi \,\,\frac{{x\,dx}}{{1 + \cos \,\,\alpha \,\,\sin \,\,x}},\;\,0 < \alpha < \pi

, is

View answer and explanation

Correct option: A

\frac{{\pi \alpha }}{{\sin \,\alpha }}

Option A (

\frac{{\pi \alpha }}{{\sin \alpha }}

) is correct because the integral is solved using properties of definite integrals and trigonometric substitutions.

879 Medium

879 of 995

Application +4 | -1

A

is the area bounded by

x+2 y^2=0

and

x+3 y^2=1

, then find the value of

3 A

View answer and explanation

Correct option: A

Option A (4) is correct because the area bounded by the curves is calculated through integration, and multiplying the result by 3 gives 4.

880 Easy

880 of 995

Application +4 | -1

{I_1} = \int \nolimits_0^{\pi /2} \,\,\sin \,\,x\,\,dx,\,\,{I_2} = \int \nolimits_0^{\pi /2} \,\,{\sin ^3}\,\,x\,\,dx

then

View answer and explanation

Correct option: A

{I_1}\,\,\, > \,\,\,{I_2}

Option A is correct because sin(x) is greater than sin³(x) in the interval [0, pi/2], therefore the integral of sin(x) is greater than the integral of sin³(x).

881 Easy

881 of 995

Application +4 | -1

The value of the integral

\int \limits_0^\pi \,\,\frac{1}{{{a^2} - \,\,2a\,\,\cos \,\,x\,\, + \,\,1}}\,\,dx\,\,(a < 1)

View answer and explanation

Correct option: A

\frac{\pi }{{1\,\, - \,\,{a^2}}}

Option A (

\frac{\pi }{{1\,\, - \,\,{a^2}}}

) is correct because after applying trigonometric identities and substitution, the integral evaluates to the given expression.

882 Easy

882 of 995

Application +4 | -1

\int_0^\pi x f(\sin x) d x

is equal to

View answer and explanation

Correct option: D

\pi \int_0^{\pi / 2} f(\cos x) d x

Option D is correct because the integral can be simplified using the property

\int_0^a f(x) dx = \int_0^a f(a-x) dx

and trigonometric identities to arrive at the given result.

883 Easy

883 of 995

Application +4 | -1

A

is the area between the curve

y = \sin x

and

x -

axis in the interval

\left[ {0,\pi /4} \right]

, then in the same interval, area between the curve

y = \cos x

and

x

-axis is

View answer and explanation

Correct option: C

1 - A

Option C (1 - A) is correct because the area under cos(x) from 0 to π/4 can be expressed as 1 - A, where A is the area under sin(x) in the same interval.

884 Easy

884 of 995

Application +4 | -1

f\,\,\left( x \right) = \tan x\,\, - \,\,{\tan ^3}x\,\, + \,\,{\tan ^5}\,\,x\,\, - ...\,\infty \,\,\mathrm{with}\,\,0\,\, < \,\,x\,\, < \,\,\frac{\pi }{4},\,\,\mathrm{then}\mathop {\,\,\int }\nolimits_0^{\pi /4} \,\,f\,\,\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{4}

Option C (

\frac{1}{4}

) is correct because the function simplifies to

\frac{{\sin \,2x}}{2}

, and its integral from 0 to

\frac{\pi}{4}

\frac{1}{4}

885 Medium

885 of 995

Application +4 | -1

f(x)=\int\limits_{0}^x \cos ^4 t d t

, then

f(x+\pi)

equals

View answer and explanation

Correct option: A

f(x)+f(\pi)

Option A is correct because by using the property of definite integrals and trigonometric identities, we can show that f(x+

\pi

) = f(x) + f(

\pi

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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

886 Easy

886 of 995

Application +4 | -1

The odd natural number

a

, such that the area of the region bounded by

y=1, y=3, x=0, x=y^a

\frac{364}{3}

, is equal to

View answer and explanation

Correct option: B

Option B (5) is correct because it satisfies the integral equation derived from the given area and limits.

887 Easy

887 of 995

Application +4 | -1

The value of

\int \nolimits_1^2 \,\,{\left[ {\,f\,\,\left\{ {g\,\,\left( x \right)} \right\}} \right]^{ - 1}}f'\left\{ {g\,\,\left( x \right)} \right\}\,\,g'\,\,\left( x \right)\,\,dx

, where

g\left( 1 \right) = g\left( 2 \right)

is equal to

View answer and explanation

Correct option: C

\mathrm{0}

Option C (

\mathrm{0}

) is correct because the integral evaluates to zero since the limits result in the same value due to g(1) = g(2).

888 Easy

888 of 995

Application +4 | -1

The value of

\int \nolimits_{{e^{ - 1}}}^{{e^2}} \,\,\left| {\,\frac{{{{\log }_e}\,\,x}}{x}\,} \right|\,\,dx

View answer and explanation

Correct option: B

\frac{5}{2}

Option B (

\frac{5}{2}

) is correct because the integral is split into two parts based on the absolute value, and then solved using substitution.

889 Easy

889 of 995

Application +4 | -1

\int \nolimits_0^{\pi /8} \,\,{\cos ^3}\,\,4\theta \,\,d\theta

is equal to

View answer and explanation

Correct option: D

\frac{1}{6}

Option D (

\frac{1}{6}

) is correct because the integral of cos³(4θ) from 0 to π/8 evaluates to 1/6.

890 Hard

890 of 995

Application +4 | -1

\int \limits_{ - \pi }^\pi \,\,{\left[ {\cos \,\,px\,\, - \,\,\sin \,\,qx} \right]^2}dx

, where

p,\;\,q

are integers is equal to

View answer and explanation

Correct option: D

2\,\pi

Option D (

2\pi

) is correct because when integrating the square of

(\cos px - \sin qx)

from

-\pi

\pi

, the cross term vanishes due to symmetry, and the integrals of

\cos^2 px

and

\sin^2 qx

each contribute

\pi

891 Easy

891 of 995

Application +4 | -1

\int\limits_{0}^{10}[x]^3\{x\} d x

is equal to:

View answer and explanation

Correct option: B

\frac{2025}{2}

Option B (

\frac{2025}{2}

) is correct because the integral is evaluated by splitting it into intervals and using the properties of greatest integer function and fractional part function.

892 Medium

892 of 995

Analyzing +4 | -1

f\,\,\left( x \right) = \int \nolimits_{{x^2}}^{{x^2} + 1} \,\,{e^{ - {t^2}}}\,dt

, then

f\,\,\left( x \right)

increases in

View answer and explanation

Correct option: D

\left( { - \,\infty ,\;\,0} \right)

Option D ((

- \,\infty ,\;\,0

)) is correct because the derivative of the function is positive when x < 0, indicating that the function is increasing in that interval.

893 Medium

893 of 995

Application +4 | -1

(\,\int \nolimits_0^a \,\,x\,dx\,)\;\,\, \le \,\,\,\left( {a + 4} \right)

, then

View answer and explanation

Correct option: B

- \,2 \le \,\,\,a\,\, \le 4

Option B (

- \\,2 \le \\,\\,a\\,\le 4

) is correct because it represents the interval where the inequality holds true after solving the quadratic inequality.

894 Medium

894 of 995

Application +4 | -1

Let

k(x)

be a continuous function satisfying the equation

\int\limits_{0}^{x^3} k(t) d t=x^{1+x^2}

, find the value of

3 k(1)

View answer and explanation

Correct option: A

Option A (2) is correct because it's derived from evaluating the limit by converting the sum to an integral and then performing integration by parts.

895 Medium

895 of 995

Application +4 | -1

Let

\mathrm{f}:(-1,1) \rightarrow \mathrm{R}

be a continuous function. If

\int\limits_{0}^{\sin x} f(t) d t=\frac{\sqrt{3}}{2} x

, then

f\left(\frac{\sqrt{3}}{2}\right)

is equal to :

View answer and explanation

Correct option: D

\sqrt{3}

Option D (

\sqrt{3}

) is correct because it is the value obtained by differentiating the given integral equation using the Leibniz rule and substituting the appropriate value of x.

896 Easy

896 of 995

Application +4 | -1

The area bounded by the curve

y = {\sec ^2}x,\;y = 0

and

\left| x \right| = \frac{\pi }{3}

View answer and explanation

Correct option: C

2\sqrt 3

sq unit

Option C (

2\sqrt 3

sq unit) is correct because it represents the definite integral of

\sec^2 x

from

-\frac{\pi}{3}

\frac{\pi}{3}

897 Easy

897 of 995

Application +4 | -1

\int \limits_{\log 2}^x \,\,\frac{1}{{\sqrt {{e^x} - \,\,1} }}\,\,dx = \frac{\pi }{6}

is equal to

View answer and explanation

Correct option: C

\log \,\,4

Option C (

\log \,4

) is correct because after solving the definite integral with the given limits and substitution, we find that x =

\log \,4

898 Easy

898 of 995

Analyzing +4 | -1

\int \nolimits_{ - 1/2}^{1/2} \,\,\cos \,\,x\,\,\log \,\,\left( {\frac{{1 + x}}{{1 - x}}} \right)\,\,dx = k\,\,.\,\,\log \,\,2

, then

k

equals

View answer and explanation

Correct option: A

\mathrm{0}

Option A (

\mathrm{0}

) is correct because the integrand is an odd function, and the integral is evaluated over a symmetric interval.

899 Easy

899 of 995

Application +4 | -1

The value of the integral

\int \nolimits_0^{\pi /2} \,\,{\sin ^5}\,\,x\,dx

View answer and explanation

Correct option: C

\frac{8}{{15}}

Option C (

\frac{8}{{15}}

) is correct because it is the result of applying Walli's formula for the definite integral of

\sin^n x

from 0 to

\pi/2

when n is odd.

900 Easy

900 of 995

Application +4 | -1

\int \,\,\frac{{{x^5}}}{{\sqrt {1 + {x^3}} }}\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{2}{9}\,\,{\left( {1 + {x^3}} \right)^{3/2}}\,\, - \,\,\frac{2}{3}\,\,{\left( {1 + {x^3}} \right)^{1/2}} + \,\,C

Option B (

\frac{2}{9}\\ {\left( {1 + {x^3}} \right)^{3/2}}\\ - \\ \frac{2}{3}\\ {\left( {1 + {x^3}} \right)^{1/2}} + \\ C

) is correct because it is the result of the integration after applying the correct substitution and simplification.

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