But for all arbitrary constants, $\smallint \,\,\sqrt {\frac{{1\,\, + \,\,\sin \,\,\theta \,\, - \,\,{{\sin }^2}\,\,\theta \,\, - \,\,{{\sin }^3}\,\,\theta }}{{2\,\,\sin \,\,\theta \,\, - \,\,1}}\,\,} d\theta \,\,{\rm{is }}\,{\rm{equal }}\,{\rm{to}}$

$\frac{1}{{2\,\,\sqrt 2 }}\,\,\sqrt {\sin \,\,\theta \,\, - \,\,\cos \,\,2\theta } \,\, + \,\,\frac{3}{4}\,\,{\log _e}\,\,\left| {\,\left( {4\,\,\sin \,\,\theta \,\, + \,\,1} \right)\,\, - \,\,\sqrt {\sin \,\,\theta \,\, - \,\,\cos \,\,2\,\,\theta } \,} \

$\frac{1}{2}\,\,\sqrt {\,\,\sin \,\,\theta \,\, + \,\,\cos \,\,2\,\theta } \,\, + \,\,\frac{3}{{4\,\,\sqrt 2 }}\,\,{\log _e}\,\,\left| {\,4\,\,\sin \,\,\theta \,\, + \,\,1\,\, - \,\,\sqrt {\sin \,\,\theta \,\, - \,\,\cos \,\,2\,\,\theta \,} } \right|$

$\smallint \frac{{{x^4} - \,\,1}}{{{x^2}{{\left( {{x^4} + {x^2} + 1} \right)}^{1/2}}}}\;dx$ is equal to

$\sqrt {\frac{{{x^4} + {x^2} + 1}}{c}} + c$

$\frac{{{x^2}}}{{\sqrt {{x^4} + {x^2} + 1} }} + c$

$x{\left( {{x^4} + {x^2} + 1} \right)^{3/2}} + c$

$\smallint \,\,\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^{1/3}}}}\,\,dx$ is equal to

$\frac{{20}}{3}\,\,{\left( {1 + {x^2}} \right)^{2/3}}\,\,\left( {2{x^2} - 3} \right) + C$

If $f\left( x \right) = \frac{{x + 2}}{{2x + 3}}$. Then, $\smallint \,{\left( {\frac{{f\left( x \right)}}{{{x^2}}}} \right)^{1/2}}dx$ is equal to $\frac{1}{{\sqrt 2 }}\,\,{\rm{g}}\left( {\frac{{1 + \sqrt {2f\left( x \right)} }}{{1 - \sqrt {2f\left( x \right)} }}} \right)\,\, - \,\,\sqrt {\frac{2}{3}} \,\,h\left( {\frac{{\sqrt {3\,f\left( x \right)} \,\, + \,\,\sqrt 2 }}{{\sqrt {3\,f\left( x \right)} \,\, - \,\,\sqrt 2 }}} \right) + \,\,c$ where

${\rm{g}}\,\left( x \right) = {\tan ^{ - 1}}x,\;\,\,h\left( x \right) = \log \,\,\left| {\,x\,} \right|$

${\rm{g}}\,\left( x \right) = \log \,\,\left| {\,x\,} \right|,\;\,\,h\left( x \right) = {\tan ^{ - 1}}\,\,x$

${\rm{g}}\left( x \right) = h\left( x \right) = {\tan ^{ - 1}}x$

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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Hard

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1 Hard

1 of 15

Analyzing +4 | -1

But for all arbitrary constants,

\int \,\,\sqrt {\frac{{1\,\, + \,\,\sin \,\,\theta \,\, - \,\,{{\sin }^2}\,\,\theta \,\, - \,\,{{\sin }^3}\,\,\theta }}{{2\,\,\sin \,\,\theta \,\, - \,\,1}}\,\,} d\theta \,\,\mathrm{is }\,\mathrm{equal }\,\mathrm{to}

View answer and explanation

Correct option: A

\frac{1}{2}\,\,\sqrt {\sin \,\,\theta \,\, - \,\,\cos \,\,2\,\theta } \,\, + \,\,\frac{3}{{4\,\,\sqrt 2 }}\,\,{\log _e}\left| {\,\left( {4\,\,\sin \,\,\theta \,\, + \,\,1} \right)\,\, + \,\,2\,\,\sqrt 2 \,\,\sqrt {\sin \,\,\theta \,\, - \,\,\cos \,\,2\,

Option A is correct because the integral simplifies to a form that can be solved using substitution and partial fractions, resulting in the given expression.

2 Hard

2 of 15

Analyzing +4 | -1

\int \frac{{{x^4} - \,\,1}}{{{x^2}{{\left( {{x^4} + {x^2} + 1} \right)}^{1/2}}}}\;dx

is equal to

View answer and explanation

Correct option: D

\frac{{\sqrt {{x^4} + {x^2} + 1} }}{x} + c

The correct answer is D (

\frac{{\sqrt {{{{x^4} + {x^2} + 1}}} }}{x} + c

) because the integral simplifies to the derivative of the given expression.

3 Hard

3 of 15

Analyzing +4 | -1

\int x^{-11}\left(1+x^4\right)^{-1 / 2} d x= -\frac{1}{10} x^{-a}\left(1+x^4\right)^{5 / 2}+\frac{1}{3} x^{-b}\left(1+x^4\right)^{3 / 2} +\frac{-1}{2} x^{-c}\left(1+x^4\right)^{1 / 2}+k

then

a+b+c

equals

View answer and explanation

Correct option: D

Option D (18) is correct because the values of a, b, and c are determined by solving the integral and comparing coefficients, and their sum is 18.

4 Hard

4 of 15

Application +4 | -1

\int \,\,\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^{1/3}}}}\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{3}{{20}}\,\,{\left( {1 + {x^2}} \right)^{2/3}}\,\,\left( {2{x^2} - 3} \right) + C

Option B (

\frac{3}{{20}}\left( {1 + {x^2}} \right)^{2/3}\left( {2{x^2} - 3} \right) + C

) is correct because it is the result of integrating the given function using substitution and integration by parts.

5 Hard

5 of 15

Analyzing +4 | -1

f\left( x \right) = \frac{{x + 2}}{{2x + 3}}

. Then,

\int \,{\left( {\frac{{f\left( x \right)}}{{{x^2}}}} \right)^{1/2}}dx

is equal to

\frac{1}{{\sqrt 2 }}\,\,\mathrm{g}\left( {\frac{{1 + \sqrt {2f\left( x \right)} }}{{1 - \sqrt {2f\left( x \right)} }}} \right)\,\, - \,\,\sqrt {\frac{2}{3}} \,\,h\left( {\frac{{\sqrt {3\,f\left( x \right)} \,\, + \,\,\sqrt 2 }}{{\sqrt {3\,f\left( x \right)} \,\, - \,\,\sqrt 2 }}} \right) + \,\,c

where

View answer and explanation

Correct option: D

\mathrm{g}\left( x \right) = \log \,\,\left| {\,x\,} \right|,\,\,\;h\left( x \right) = \log \,\,\left| {\,x\,} \right|

Option C (

\frac{{{5^{{5^{{5^x}}}}}}}{{{{\left( {\log \,5} \right)}^3}}}\\ + \\,C

) is correct because the integral is solved using repeated substitution and the formula for integrating exponential functions.

6 Hard

6 of 15

Analyzing +4 | -1

\int \limits_{ - 1}^1 \,\,\frac{{{x^3}\,\, + \,\,\left| {\,x\,} \right|\,\, + \,\,1}}{{{x^2}\,\, + \,\,2\left| {\,x\,} \right|\,\, + \,1}}\,\,dx

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the integral evaluates to 2ln2, which is not among the given options.

7 Hard

7 of 15

Analyzing +4 | -1

\int \,\,\frac{{{x^2}\,\, - \,\,1}}{{x\,\sqrt {\left( {{x^2}\,\, + \,\,\alpha \;x\,\, + \,\,1} \right)\,\,\left( {{x^2}\,\, + \,\,\beta \;x\,\, + \,\,1} \right)} }}\,\,dx

is equal to

View answer and explanation

Correct option: A

\log \,\,\left\{ {\,\frac{{\sqrt {{x^2} + \,\,\alpha \;x\,\, + 1} \,\, + \,\,\sqrt {{x^2} + \,\,\beta \;x\,\, + 1} }}{{\sqrt x }}\,} \right\}\,\, + \,\,C

Option A (

\log \,\,\left\{ {\,\frac{{\sqrt {{{{x^2}} + \,\,\alpha \;x\,\, + 1}} \,\, + \,\,\sqrt {{{{x^2}} + \,\,\beta \;x\,\, + 1}} }}{{\sqrt x }}}\} \right\}\,\, + \,\,C

) is correct because the integral is solved by dividing by x and completing the square.

8 Hard

8 of 15

Application +4 | -1

\int\left(x^x\right)^x\left(2 x \log _e x+x\right) d x

is equal to

View answer and explanation

Correct option: B

\left(x^x\right)^x+C

Option B is correct because the integral simplifies to the derivative of (x^x)^x, making the integration straightforward.

9 Hard

9 of 15

Analyzing +4 | -1

\int \frac{d x}{(1+x) \sqrt{1+x-x^2}}

equals

View answer and explanation

Correct option: B

-2 \tan^{-1} \left(\frac{\sqrt{1+x-x^2}+1}{x}+1\right)+c

Option B (

-2 \tan^{-1} (\frac{{\sqrt{{1+x-x^2}}+1}}{{x}}+1)+c

) is correct because after applying Euler's substitution and simplifying, the integral evaluates to this form.

10 Hard

10 of 15

Understanding +4 | -1

\ell^{\mathrm{r}}

means

(\log \log \log \ldots.) \mathrm{x}

, the

\log

being repeated

\mathrm{r}

times, then

\int\left\{\mathrm{x} \ell(\mathrm{x}) \ell^2(\mathrm{x}) \ell^3(\mathrm{x}) \ldots \ell^{\mathrm{r}}(\mathrm{x})\right\}^{-1} \mathrm{dx}

is equal to -

View answer and explanation

Correct option: A

\ell^{r+1}(\mathrm{x})+\mathrm{c}

Option A (

\ell^{r+1}(\mathrm{x})+\mathrm{c}

) is correct because the integral simplifies to the (r+1)-th iterated logarithm of x plus a constant, through a series of substitutions.

11 Hard

11 of 15

Analyzing +4 | -1

Let

I(x)=\int \frac{x^{2}\left(x \sec ^{2}+\tan x\right)}{(x \tan x+1)^{2}} d x

. If

I(0)=0

, then

I\left(\frac{\pi}{4}\right)

is equal to

View answer and explanation

Correct option: C

\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}

Option C (

\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}

) is correct because it is the result of evaluating the definite integral with the given initial condition.

12 Hard

12 of 15

Understanding +4 | -1

\int\left(\frac{f(x) g^{\prime}(x)-f^{\prime}(x) g(x)}{f(x) g(x)}\right)(\log (g(x))-\log (f(x))) d x

is equal to:

View answer and explanation

Correct option: C

\frac{1}{2}\left(\log \left(\frac{g(x)}{f(x)}\right)\right)^2+C

Option C (

\frac{{1}}{{2}}\left(\log \left(\frac{{g(x)}}{{f(x)}}\right)\right)^2+C

) is correct because it is the result of integrating the given expression after a suitable substitution.

13 Hard

13 of 15

Application +4 | -1

The value of

\int \limits_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

Option D (None of these) is correct because the integral evaluates to 0 due to the given condition g(1) = g(2).

14 Hard

14 of 15

Analyzing +4 | -1

\int \frac{e^{\sec x} \sin [x+(\pi / 4)]}{\cos x(1-\sin x)} d x\left(0 < x < \frac{\pi}{4}\right)

is equal to

View answer and explanation

Correct option: B

\frac{1}{\sqrt{2}} e^{\sec x}(\sec x+\tan x)+c

Option B is correct because after simplification and integration, the integral evaluates to

\frac{1}{\sqrt{2}} e^{\sec x}(\sec x+\tan x)+c

15 Hard

15 of 15

Application +4 | -1

Let

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

be a function satisfying

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

with

f\left( 0 \right) = 1

and

g\,\,\left( x \right)

be a function that satisfies

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

. Then, the value of the integral

\int \limits_0^1 \,f\left( x \right)\;\,\,g\,\,\left( x \right)\;\,dx

, is

View answer and explanation

Correct option: D

e\,\, - \,\,\frac{{{e^2}}}{2}\,\, - \,\,\frac{3}{2}

Option D (

e\,\, - \,\,\frac{{{{{e^2}}}}}{2}\,\, - \,\,\frac{3}{2}

) is correct because it is the result of evaluating the definite integral after finding the functions f(x) and g(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

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