Indefinite Integration — Page 4 | JEE Mathematics

Q31

The integral

\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}}

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

A

{1 \over 2}{\left( {{{x - 3} \over {x + 4}}} \right)^{{3 \over 7}}} + C

B

{\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}} + C

C

- {1 \over {13}}{\left( {{{x - 3} \over {x + 4}}} \right)^{{{13} \over 7}}} + C

D -

{\left( {{{x - 3} \over {x + 4}}} \right)^{-{1 \over 7}}} + C

Correct Answer

Option B

Solution

\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}}

=

\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}}

Put

{{{x - 3} \over {x + 4}}}

= t $\Rightarrow$

\left\{ {{{\left( {x + 4} \right) - \left( {x - 3} \right)} \over {{{\left( {x + 4} \right)}^2}}}} \right\}dx

= dt $\Rightarrow$

{{dx} \over {{{\left( {x + 4} \right)}^2}}} = {{dt} \over 7}

=

{1 \over 7}\int {{{dt} \over {{{\left( t \right)}^{{6 \over 7}}}}}}

=

{1 \over 7}\left( {{{{t^{{1 \over 7}}}} \over {{1 \over 7}}}} \right)

+ C =

{\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}}

+ C

Q32

If ƒ'(x) = tan–1(secx + tanx),

- {\pi \over 2} < x < {\pi \over 2}

, and ƒ(0) = 0, then ƒ(1) is equal to :

A

{1 \over 4}

B

{{\pi - 1} \over 4}

C

{{\pi + 1} \over 4}

D

{{\pi + 2} \over 4}

Correct Answer

Option C

Solution

ƒ'(x) = tan–1(secx + tanx) $\Rightarrow$ ƒ'(x) =

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

=

{\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)

=

{\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)

As

- {\pi \over 2} < x < {\pi \over 2}

$\Rightarrow$

0 < {\pi \over 4} + {x \over 2} < {\pi \over 2}

$\therefore$ ƒ'(x) =

{\pi \over 4} + {x \over 2}

$\Rightarrow$ f(x) =

{\pi \over 4}x + {{{x^2}} \over 4} + c

$\because$ ƒ(0) = 0 $\Rightarrow$ c = 0 $\therefore$ f(x) =

{\pi \over 4}x + {{{x^2}} \over 4}

$\Rightarrow$ f(1) =

{{{\pi + 1} \over 4}}

Q33

If

\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c

, where c is a constant of integration, then the ordered pair (a, b) is equal to :

A (-1, 3)

B (1, 3)

C (1, -3)

D (3, 1)

Correct Answer

Option B

Solution

Given

\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx

Write sin2x = 1 + sin2x - 1 =

\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx

=

\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx

=

\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx

=

\int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} \right)}^2}} }}} dx

put sin x + cos x = t $\Rightarrow$ (cos x – sin x) dx = dt =

\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}}

=

{\sin ^{ - 1}}\left( {{t \over 3}} \right)

+ C =

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

$\therefore$ a = 1 and b = 3

Q34

For

I(x)=\int \dfrac{\sec ^{2} x-2022}{\sin ^{2022} x} d x

, if

I\left(\dfrac{\pi}{4}\right)=2^{1011}

, then

A

3^{1010} I\left(\dfrac{\pi}{3}\right)-I\left(\dfrac{\pi}{6}\right)=0

B

3^{1010} I\left(\dfrac{\pi}{6}\right)-I\left(\dfrac{\pi}{3}\right)=0

C

3^{1011} I\left(\dfrac{\pi}{3}\right)-I\left(\dfrac{\pi}{6}\right)=0

D

3^{1011} I\left(\dfrac{\pi}{6}\right)-I\left(\dfrac{\pi}{3}\right)=0

Correct Answer

Option A

Solution

Given,

I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx}

= \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }

= \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }

= {1 \over {{{\sin }^{2022}}x}}\,.\,\tan x - \int {\left( {{{ - 2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,\tan x} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} }

= {{\tan x} \over {{{\sin }^{2022}}x}} + \int {\left( {{{2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,{{\sin x} \over {\cos x}}} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} }

= {{\tan x} \over {{{\sin }^{2022}}x}} + \int {{{2022} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} }

= {{\tan x} \over {{{\sin }^{2022}}x}} + C

Given,

I\left( {{\pi \over 4}} \right) = {2^{1011}}

$\therefore$

I\left( {{\pi \over 4}} \right) = {{\tan \left( {{\pi \over 4}} \right)} \over {{{\left( {\sin {\pi \over 4}} \right)}^{2022}}}} + C

\Rightarrow {2^{1011}} = {1 \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^{2022}}}} + C

\Rightarrow C = {2^{1011}} - {2^{1011}} = 0

$\therefore$

I(x) = {{\tan x} \over {{{\sin }^{2022}}x}}

$\therefore$

I\left( {{\pi \over 3}} \right) = {{\tan {\pi \over 3}} \over {{{\left( {\sin {\pi \over 3}} \right)}^{2022}}}} = {{\sqrt 3 } \over {{{\left( {{{\sqrt 3 } \over 2}} \right)}^{2022}}}}

I\left( {{\pi \over 6}} \right) = {{{1 \over {\sqrt 3 }}} \over {{{\left( {{1 \over 2}} \right)}^{2022}}}} = {1 \over {\sqrt 3 }} \times {(2)^{2022}}

From option (A),

{3^{1010}}\,.\,I\left( {{\pi \over 3}} \right) - I\left( {{\pi \over 6}} \right)

= {3^{1010}}\,.\,\sqrt 3 \,.\,{\left( {{2 \over {\sqrt 3 }}} \right)^{2022}} - {{{{(2)}^{2022}}} \over {\sqrt 3 }}

= {3^{1010}}\,.\,\sqrt 3 \times {{{2^{2022}}} \over {{3^{1011}}}} - {{{2^{2022}}} \over {\sqrt 3 }}

= {{{2^{2022}}} \over {\sqrt 3 }} - {{{2^{2022}}} \over {\sqrt 3 }} = 0

Q35

The value of the integral

\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta

is :

A

{1 \over {18}}{\left[ {9 - 2{{\cos }^6}\theta - 3{{\cos }^4}\theta - 6{{\cos }^2}\theta } \right]^{{3 \over 2}}} + c

B

{1 \over {18}}{\left[ {11 - 18{{\sin }^2}\theta + 9{{\sin }^4}\theta - 2{{\sin }^6}\theta } \right]^{{3 \over 2}}} + c

C

{1 \over {18}}{\left[ {11 - 18{{\cos }^2}\theta + 9{{\cos }^4}\theta - 2{{\cos }^6}\theta } \right]^{{3 \over 2}}} + c

D

{1 \over {18}}{\left[ {9 - 2{{\sin }^6}\theta - 3{{\sin }^4}\theta - 6{{\sin }^2}\theta } \right]^{{3 \over 2}}} + c

Correct Answer

Option C

Solution

\int {{{2{{\sin }^2}\theta \cos \theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {2{{\sin }^2}\theta }}d\theta }

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

= \int {({t^6} + {t^4} + {t^2})\sqrt {2{t^4} + 3{t^2} + 6} \,dt}

= \int {({t^5} + {t^3} + t)\sqrt {2{t^6} + 3{t^4} + 6{t^2}} dt}

Let

2{t^6} + 3{t^4} + 6{t^2} = z

12({t^5} + {t^3} + t)dt = dz

= {1 \over {12}}\int {\sqrt z dz = {1 \over {18}}{z^{3/2}} + c}

= {1 \over {18}}[{(2{\sin ^6}\theta + 3{\sin ^4}\theta + 6{\sin ^2}\theta )^{3/2}} + C

= {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{(1 - {\cos ^2}\theta )^2} + 3 - 3{\cos ^2}\theta + 6)]^{3/2}} + C

= {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{\cos ^4}\theta - 7{\cos ^2}\theta + 11)]^{3/2}} + C

= {1 \over {18}}{[ - 2{\cos ^6}\theta + 9{\cos ^4}\theta - 18{\cos ^2}\theta + 11]^{3/2}} + C

Q36

The integral

\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx

, x > 0, is equal to : (where c is a constant of integration)

A

{\log _e}\sqrt {{x^2} + 5x - 7} + c

B

4{\log _e}|{x^2} + 5x - 7| + c

C

{1 \over 4}{\log _e}|{x^2} + 5x - 7| + c

D

{\log _e}|{x^2} + 5x - 7| + c

Correct Answer

Option B

Solution

\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}dx}

= \int {{{8{x^3} + 5(4{x^2})} \over {{x^4} + 5{x^3} - 7{x^2}}}}

= \int {{{8{x^3} + 20{x^2}} \over {{x^4} + 5{x^3} - 7{x^2}}}}

= \int {{{8x + 20} \over {{x^2} + 5x - 7}}}

= \int {{{4(2x + 5)} \over {{x^2} + 5x - 7}}}

Let

\left\{ \begin{array}{ll}{x^2} + 5x - 7 = t \\ (2x + 5)dx = dt \end{array} \right\}

= \int {{{4dt} \over t}}

= 4\ln \left| t \right| + C

= 4\ln \left| {({x^2} + 5x - 7)} \right| + C

Here C is integral constant.

Q37

The integral

\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx

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

A

{1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c

B

{1 \over 2}\cos \sqrt {{{(2x + 1)}^2} + 5} + c

C

{1 \over 2}\cos \sqrt {{{(2x - 1)}^2} + 5} + c

D

{1 \over 2}\sin \sqrt {{{(2x + 1)}^2} + 5} + c

Correct Answer

Option A

Solution

\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {{{(2x - 1)}^2} + 5} }}} dx

{(2x - 1)^2} + 5 = {t^2}

2(2x - 1)2dx = 2t\,dt

2\sqrt {{t^2} - 5} dx = t\,dt

So,

\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c}

= {1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c

Q38

The integral

\int {{1 \over {\sqrt[4]{{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx

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

A

{3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{1 \over 4}}} + C

B

{3 \over 4}{\left( {{{x + 2} \over {x - 1}}} \right)^{{5 \over 4}}} + C

C

{4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C

D

{4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{5 \over 4}}} + C

Correct Answer

Option C

Solution

\int {{{dx} \over {{{(x - 1)}^{3/4}}{{(x + 2)}^{5/4}}}}}

= \int {{{dx} \over {{{\left( {{{x + 2} \over {x - 1}}} \right)}^{5/4}}.\,{{(x - 1)}^2}}}}

put

{{x + 2} \over {x - 1}} = t

= - {1 \over 3}\int {{{dt} \over {{t^{5/4}}}}}

= {4 \over 3}.{1 \over {{t^{1/4}}}} + C

= {4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{1/4}} + C

Q39

If

\int {{{({x^2} + 1){e^x}} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + C}

, where C is a constant, then

{{{d^3}f} \over {d{x^3}}}

at x = 1 is equal to :

A

- {3 \over 4}

B

{3 \over 4}

C

- {3 \over 2}

D

{3 \over 2}

Correct Answer

Option B

Solution

I = \int {{{{e^x}({x^2} + 1)} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + c}

= \int {{{{e^x}({x^2} - 1 + 1 + 1)} \over {{{(x + 1)}^2}}}dx}

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

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

$\therefore$

f(x) = {{x - 1} \over {x + 1}}

f(x) = 1 - {2 \over {x + 1}}

f'(x) = 2{\left( {{1 \over {x + 1}}} \right)^2}

f''(x) = - 4{\left( {{1 \over {x + 1}}} \right)^3}

f'''(x) = {{12} \over {{{(x + 1)}^4}}}

for

x = 1

f'''(1) = {{12} \over {{2^4}}} = {{12} \over {16}} = {3 \over 4}

Q40

If

\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c}

,

g(1) = 0

, then

g\left( {{1 \over 2}} \right)

is equal to :

A

{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) + {\pi \over 3}

B

{\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) + {\pi \over 3}

C

{\log _e}\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right) - {\pi \over 3}

D

{1 \over 2}{\log _e}\left( {{{\sqrt 3 - 1} \over {\sqrt 3 + 1}}} \right) - {\pi \over 6}

Correct Answer

Option A

Solution

Given,

\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c}

,

g(1) = 0

Let I = $\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx}$ = $\int {{1 \over x}\sqrt {{{\left( {1 - x} \right)\left( {1 + x} \right)} \over {\left( {1 + x} \right)\left( {1 + x} \right)}}} } dx$ $=\int \dfrac{1-x}{x \sqrt{1-x^2}} d x$ $=\int \dfrac{1}{x \sqrt{1-x^2}} d x-\int \dfrac{1}{\sqrt{1-x^2}} d x$ Put $x=\dfrac{1}{t}$ $d x=-\dfrac{1}{t^2}$ $=\int \dfrac{\dfrac{-1}{t^2}}{\dfrac{1}{t} \sqrt{1-\dfrac{1}{t^2}}} d t-\sin ^{-1}(x)+C_1$ $=\int \dfrac{-d t}{\sqrt{t^2-1}}-\sin ^{-1}(x)+C_1$ $=-\ln \left|t+\sqrt{t^2-1}\right|-\sin ^{-1}(x)+C_1$ $=-\ln \left|\dfrac{1}{\mathrm{x}}+\sqrt{\dfrac{1}{\mathrm{x}^2}-1}\right|-\sin ^{-1}(\mathrm{x})+\mathrm{C}_1$ [Putting values of t] $=-\ln \left|\dfrac{1}{\mathrm{x}}+\sqrt{\dfrac{1}{\mathrm{x}^2}-1}\right|-\left(\dfrac{\pi}{2}-\cos ^{-1} \mathrm{x}\right)+\mathrm{C}_1$ $=-\ln \left|\dfrac{1}{\mathrm{x}}+\sqrt{\dfrac{1}{\mathrm{x}^2}-1}\right|+\cos ^{-1}(\mathrm{x})-\dfrac{\pi}{2}+\mathrm{C}_1$ $\therefore$ $g(x)=\cos ^{-1}(x)-\ell n\left|\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}-1}\right|$ So, $g(1)=\cos ^{-1}(1)-\ell n|1|=0$ $g\left(\dfrac{1}{2}\right)=\cos ^{-1}\left(\dfrac{1}{2}\right)-\ell n|2+\sqrt{3}|$ $= {\pi \over 3} + \ln \left( {{1 \over {2 + \sqrt 3 }}} \right)$ $=\dfrac{\pi}{3}+\ln \left(\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right)$