Indefinite Integration — Page 3 | JEE Mathematics

Q21

\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx}

is equal to (where c is a constant of integration)

A 2x + sinx + 2sin2x + c

B x + 2sinx + sin2x + c

C x + 2sinx + 2sin2x + c

D 2x + sinx + sin2x + c

Correct Answer

Option B

Solution

\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx}

=

\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx

=

\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx

=

\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx

Use sin2x = 2sinxcosx and sin3x = 3sinx – 4sin3x =

\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx

=

\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx

=

\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx

=

\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx

= x + sin2x + 2sinx + C

Q22

The integral

\int \,

cos(loge x) dx is equal to : (where C is a constant of integration)

A

{x \over 2}

[sin(loge x)

-

cos(loge x)] + C

B x[cos(loge x) + sin(loge x)] + C

C

{x \over 2}

[cos(loge x) + sin(loge x)] + C

D x[cos(loge x)

-

sin(loge x)] + C

Correct Answer

Option C

Solution

{\rm I} = \int {\cos \left( {\ell nx} \right)} dx

{\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx}

{\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]

{\rm I} = {x \over 2}\left[ {\sin \left( {\ell nx} \right) + \cos \left( {\ell nx} \right)} \right] + C

Q23

\text{ The integral } \int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x \text{ is equal to }

A

\dfrac{1}{2} \log _{e}\left|\dfrac{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)}{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}\right|+C

B

\dfrac{1}{2} \log _{e}\left|\dfrac{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)}\right|+C

C

\log _{e}\left|\dfrac{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{6}\right)}{\tan \left(\dfrac{x}{2}+\dfrac{\pi}{12}\right)}\right|+C

D

\dfrac{1}{2} \log _{e}\left|\dfrac{\tan \left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)}{\tan \left(\dfrac{x}{2}-\dfrac{\pi}{6}\right)}\right|+C

Correct Answer

Option A

Solution

= \int {{{\left( {1 - {1 \over {\sqrt 3 }}} \right)(\cos x - \sin x)} \over {\left( {1 + {2 \over {\sqrt 3 }}\sin 2x} \right)}}dx}

= \int {{{\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right)\sqrt 2 \sin \left( {{\pi \over 4} - x} \right)} \over {\left( {{2 \over {\sqrt 3 }}} \right)\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx}

= \int {{{{{(\sqrt 3 - 1)} \over {\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx}

= \int {{{{{\sqrt 3 - 1} \over {2\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}

= {1 \over 2}\int {{{2\sin {\pi \over {12}}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}

= {1 \over 2}\int {{{\cos \left( {{\pi \over 6} - x} \right) - \cos \left( {{\pi \over 3} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx}

= {1 \over 2}\left[ {\int {{\mathop{\rm cosec}\nolimits} \left( {{\pi \over 6} + x} \right)dx - \int {\sec \left( {{\pi \over 6} - x} \right)dx} } } \right]

= {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \int {\cos ec\left( {{\pi \over 3} - x} \right)dx} } \right]

= {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \ln \left| {{\pi \over 6} + {x \over 2}} \right|} \right] + C

= {1 \over 2}\ln \left| {{{\tan \left( {{\pi \over 2} + {x \over 2}} \right)} \over {\tan \left( {{\pi \over 6} + {x \over 2}} \right)}}} \right| + C

Q24

If

\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta

= A

{\log _e}\left| {B\left( \theta \right)} \right| + C

, where C is a constant of integration, then

{{{B\left( \theta \right)} \over A}}

can be :

A

{{2\sin \theta + 1} \over {5\left( {\sin \theta + 3} \right)}}

B

{{2\sin \theta + 1} \over {\sin \theta + 3}}

C

{{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}

D

{{5\left( {\sin \theta + 3} \right)} \over {2\sin \theta + 1}}

Correct Answer

Option C

Solution

\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta

=

\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}}

=

\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}}

Let sin $\theta$ = t $\Rightarrow$ cos $\theta$ d $\theta$ = dt =

\int {{{dt} \over {3 + 7t + 2{t^2}}}}

=

\int {{{dt} \over {\left( {2t + 1} \right)\left( {t + 3} \right)}}}

=

{1 \over 5}\int {\left( {{2 \over {2t + 1}} - {1 \over {t + 3}}} \right)dt}

=

{1 \over 5}\ln \left| {{{2t + 1} \over {t + 3}}} \right|

+ C =

{1 \over 5}\ln \left| {{{2\sin \theta + 1} \over {\sin \theta + 3}}} \right|

+ C $\therefore$ A =

{{1 \over 5}}

and B( $\theta$ ) =

{{{2\sin \theta + 1} \over {\sin \theta + 3}}}

$\therefore$

{{{B\left( \theta \right)} \over A}}

=

{{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}

Q25

If

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

=

g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c

where c is a constant of integration, then g(0) is equal to :

A 1

B 2

C e

D e2

Correct Answer

Option B

Solution

I =

\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx}

=

\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx

=

\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx

+

\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

=

\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx

+ \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

=

\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx

+

\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx}

Let

{{e^x} + {e^{ - x}} + x}

= t $\Rightarrow$

{\left( {{e^x} - {e^{ - x}} + 1} \right)}

dx = dt and let

{{e^x} + {e^{ - x}}}

= u $\Rightarrow$

{\left( {{e^x} - {e^{ - x}}} \right)dx}

= du $\therefore$ I =

\int {{e^t}} dt + \int {{e^u}} du

=

{{e^t}}

+

{{e^u}}

+ C =

{e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}

+

{{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}

+ C =

{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)

+ C $\therefore$ g(x) = ex + 1 $\Rightarrow$ g(0) = 2

Q26

The integral

\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx}

is equal to (where C is a constant of integration):

A

\sec x - {{x\tan x} \over {x\sin x + \cos x}} + C

B

\sec x + {{x\tan x} \over {x\sin x + \cos x}} + C

C

\tan x - {{x\sec x} \over {x\sin x + \cos x}} + C

D

\tan x + {{x\sec x} \over {x\sin x + \cos x}} + C

Correct Answer

Option C

Solution

{\int {\left( {{x \over {x\sin x + \cos x}}} \right)} ^2}dx

= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}}

=

{x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \left( {{1 \over {x\sin x + \cos x}}} \right)dx

= - {{x\sec x} \over {x\sin x + \cos x}} + \int {{{\sec }^2}x\,dx}

= - {{x\sec x} \over {x\sin x + \cos x}} + \tan x + C

Q27

Let

f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)}

. Then f(3) – f(1) is eqaul to :

A

- {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}

B

{\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}

C

- {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}

D

{\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}

Correct Answer

Option B

Solution

\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)

Put x = tan2 $\theta$ $\Rightarrow$ 2xdx = 2tan $\theta$ sec2 $\theta$ d $\theta$

I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } }

= \theta - {{\sin 2\theta } \over 2} + c

\Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c

f(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c

Now

f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}

= {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}

= {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}

Q28

If

\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C

where C is a constant of integration, then the ordered pair (

\lambda

, ƒ(

\theta

)) is equal to :

A (–1, 1 – tan

\theta

)

B (1, 1 + tan

\theta

)

C (–1, 1 + tan

\theta

)

D (1, 1 – tan

\theta

)

Correct Answer

Option C

Solution

I =

\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}}

=

\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}}

=

\int {{{\left( {1 - {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \over {{{\left( {1 + {{\tan }}\theta } \right)}^2}}}}

tan $\theta$ = t $\Rightarrow$ sec2 $\theta$ d $\theta$ = dt =

\int {{{\left( {1 - {t^2}} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}}

=

\int {{{\left( {1 - t} \right)\left( {1 + t} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}}

=

\int {{{\left( {1 - t} \right)} \over {\left( {1 + t} \right)}}} dt

=

\int {\left( {{1 \over {\left( {1 + t} \right)}} - {t \over {\left( {1 + t} \right)}}} \right)} dt

=

{\log _e}\left| {1 + t} \right|

-

\int {\left( {{{1 + t} \over {\left( {1 + t} \right)}} - {1 \over {\left( {1 + t} \right)}}} \right)} dt

=

{\log _e}\left| {1 + t} \right|

- t +

{\log _e}\left| {1 + t} \right|

+ C = 2

{\log _e}\left| {1 + t} \right|

- t + C = 2

{\log _e}\left| {1 + \tan \theta } \right|

- tan $\theta$ + C $\therefore$ $\lambda$ = -1 and ƒ( $\theta$ ) = 1 + tan $\theta$

Q29

If

\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx

= A(x)

{\tan ^{ - 1}}\left( {\sqrt x } \right)

+ B(x) + C, where C is a constant of integration, then the ordered pair (A(x), B(x)) can be :

A (x + 1, -

{\sqrt x }

)

B (x + 1,

{\sqrt x }

)

C (x - 1, -

{\sqrt x }

)

D (x - 1,

{\sqrt x }

)

Correct Answer

Option A

Solution

Given, I =

\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx

Let

{\sin ^{ - 1}}\left( {{{\sqrt x } \over {\sqrt {1 + x} }}} \right)

= $\theta$ $\Rightarrow$

{{{\sqrt x } \over {\sqrt {1 + x} }} = \sin \theta }

$\Rightarrow$ tan $\theta$ =

{{{\sqrt x } \over 1}}

$\Rightarrow$ $\theta$ =

{\tan ^{ - 1}}\left( {\sqrt x } \right)

$\therefore$ I =

\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right)dx}

=

\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right).1dx}

Applying integration by parts, I =

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{1 \over {1 + x}}{1 \over {2\sqrt x }}xdx}

Let

{\sqrt x }

= t $\Rightarrow$ x = t2 $\Rightarrow$ dx = 2tdt $\therefore$ I =

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{{t^2}} \over {\left( {1 + {t^2}} \right)\left( {2t} \right)}}2tdt}

=

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{\left( {{t^2} + 1} \right) - 1} \over {\left( {1 + {t^2}} \right)}}dt}

=

{\tan ^{ - 1}}\left( {\sqrt x } \right).x - t + {\tan ^{-1}}t + c

=

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

$\therefore$ A(x) = x + 1, B(x) = –

{\sqrt x }

Q30

If

\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c

where c is a constant of integration, then

\lambda f\left( {{\pi \over 3}} \right)

is equal to

A

{9 \over 8}

B 2

C -2

D

-{9 \over 8}

Correct Answer

Option C

Solution

Given I =

\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}}

Let sin x = t $\Rightarrow$ cos xdx = dt $\therefore$ I =

\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}}

I =

\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6}}} + 1} \right)}^{2/3}}}}}

Let

{{1 \over {{t^6}}} + 1}

= z3 $\Rightarrow$

- {6 \over {{t^7}}}dt = 3{z^2}dz

$\Rightarrow$

{{dt} \over {{t^7}}} = {{{z^2}} \over { - 2}}dz

So I =

- \int {{{{z^2}dz} \over {2{{\left( {{z^3}} \right)}^{2/3}}}}}

I =

- \int {{{dz} \over 2}}

I =

- {z \over 2} + C

I =

- {1 \over 2}{\left( {1 + {1 \over {{t^6}}}} \right)^{{1 \over 3}}} + C

I =

- {1 \over 2}{\left( {1 + {1 \over {{{\sin }^6}x}}} \right)^{{1 \over 3}}} + C

I =

- {1 \over {2{{\sin }^2}x}}{\left( {{{\sin }^6}x + 1} \right)^{{1 \over 3}}} + C

$\therefore$ $\lambda$ = 3 and f(x) =

- {1 \over 2}

cosec2 x Then

\lambda f\left( {{\pi \over 3}} \right)

= 3 $\times$

- {1 \over 2}

cosec2

{\pi \over 3}

= 3 $\times$

- {1 \over 2}

$\times$

{4 \over 3}

= -2