Indefinite Integration — Page 2 | JEE Mathematics

Q11

The integral

\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx

is equal to :

A

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

B

{{ - {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

C

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

D

{{ {x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

Correct Answer

Option D

Solution

\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx

Dividing by

{x^{15}}

in numerator and denominator

\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}}

Substitute

1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t

\Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt

\Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt

This gives,

\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}}

= \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C

= {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C

= {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C

Q12

The integral

\int {\sqrt {1 + 2\cot x(\cos ecx + \cot x)\,} \,\,dx}

\left( {0 < x < {\pi \over 2}} \right)

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

A 4 log(sin

{x \over 2}

) + C

B 2 log(sin

{x \over 2}

) + C

C 2 log(cos

{x \over 2}

) + C

D 4 log(cos

{x \over 2}

) + C

Correct Answer

Option B

Solution

Let, I =

\int {\sqrt {1 + 2\cot x\cos ec + 2{{\cot }^2}x} .dx}

$\Rightarrow$ I =

\int {\sqrt {{{{{\sin }^2}x + 2\cos x + 2{{\cos }^2}x} \over {{{\sin }^2}x}}} .dx}

$\Rightarrow$ I =

\int {\sqrt {{{1 + 2\cos x + {{\cos }^2}x} \over {\sin x}}} .dx}

$\Rightarrow$ I =

\int {\left| {{{1 + \cos x} \over {\sin x}}} \right|dx}

$\Rightarrow$ I =

\int {\left| {\cos ec\,x + \cot x} \right|.dx}

$\Rightarrow$ I =

\log \left| {\cos ec\,x - \cot x} \right| + \log \left| {\sin x} \right| + C

$\Rightarrow$ I =

\log \left| {1 - \cos x} \right| + C

$\Rightarrow$ I =

\log \left| {2{{\sin }^2}{x \over 2}} \right| + C

$\Rightarrow$ I =

\log \left| {{{\sin }^2}{x \over 2}} \right| + \log 2+ C

$\Rightarrow$ I = 2

\log \left| {{{\sin }}{x \over 2}} \right| + C_1

Q13

If

f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,

(x

\in

R

-

{1,

-

2}), then

\int f \left( x \right)dx

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

A 12 loge | 1

-

x | + 3x + C

B

-

12 loge | 1

-

x |

-

3x + C

C 12 loge | 1

-

x |

-

3x + C

D

-

12 loge | 1

-

x | + 3x + C

Correct Answer

Option B

Solution

Let,

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

= t $\Rightarrow$ x - 4 = t(x+2) $\Rightarrow$ x (1 -t) = 2(t+2) $\Rightarrow$ x =

{{2(t + 2)} \over {1 - t}}

$\therefore$ f(t) = 2(

{{2(t + 2)} \over {1 - t}}

) + 1 =

{{4t + 8} \over {1 - t}} + 1

=

{{3t + 9} \over {1 - t}}

=

{{3(t + 3)} \over {1 - t}}

=

{{3(t - 1 + 4)} \over {1 - t}}

= - 3 +

12 \over 1 - t

$\therefore$ f(x) = - 3 +

12 \over 1 - x

$\therefore$

\int {f(x)dx}

=

\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx}

=

-12{\log _e}|1 - x|

- 3x + C

Q14

The integral

\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx

is equal to

A

{{ - 1} \over {1 + {{\cot }^3}x}} + C

B

{1 \over {3\left( {1 + {{\tan }^3}x} \right)}} + C

C

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

D

{1 \over {1 + {{\cot }^3}x}} + C

Correct Answer

Option C

Solution

Given,

\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx

= \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]}^2}}}} \,dx

= \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx

= \int {{{{{\sin }^2}x\,\,{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,\,dx

= \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {1 + {{\tan }^3}} \right)}^2}}}} \,\,dx

= \int {{{{{\tan }^2}x.{{\sec }^2}x} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,dx

[ Let

\,\,\,\,\,\,{\tan ^3}x = t

\,\,\,\,\,\,

\Rightarrow 3{\tan ^2}x\,{\sec ^2}x\,\,dx = dt

]

= {1 \over 3}\int {{{3{{\tan }^2}x\,\,{{\sec }^2}x\,\,dx} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}}

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

= {1 \over 3} \times {{{{\left( {1 + t} \right)}^{ - 2 + 1}}} \over { - 2 + 1}} + c

= {1 \over 3} \times {{ - 1} \over {\left( {1 + t} \right)}} + c

= {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c

Q15

Let

a \in \left( {0,{\pi \over 2}} \right)

be fixed. If the integral

\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx

= A(x) cos 2

\alpha

+ B(x) sin 2

\alpha

+ C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :

A

x - \alpha

and

{\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|

B

x + \alpha

and

{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|

C

x + \alpha

and

{\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|

D

x - \alpha

and

{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|

Correct Answer

Option D

Solution

To solve the given integral, first we simplify the expression in the integral as follows :

\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx

Simplifying further, this becomes :

\int {{{\sin x\cos \alpha + \sin \alpha \cos x} \over {\sin x\cos \alpha - \sin \alpha \cos x}}} dx

By using the formula for sin(a + b) = sin a cos b + cos a sin b and sin(a - b) = sin a cos b - cos a sin b, we get :

\int {{{\sin(x + \alpha)}} \over {\sin(x - \alpha)}} dx

Now, let's use the substitution method.

Let t = x - α, therefore x = y + α and dx = dt.

Substituting these values into the integral, we get : =

\int {{{\sin(t + 2\alpha)}} \over {\sin y}} dy

\begin{aligned} & =\int \frac{\sin t \cos 2 \alpha+\sin 2 \alpha \cos t}{\sin t} d t \\\\ & =\int\left(\cos 2 \alpha+\sin 2 \alpha \frac{\cos t}{\sin t}\right) d t \end{aligned}

\begin{aligned} & =t(\cos 2 \alpha)+(\sin 2 \alpha) \log _e|\sin t|+C \\\\ & =(x-\alpha) \cos 2 \alpha+(\sin 2 \alpha) \log _e|\sin (x-\alpha)|+C \\\\ & =A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C \text{ (given) } \end{aligned}

Now on comparing, we get

A(x)=x-\alpha \text{ and } B(x)=\log _e|\sin (x-\alpha)|

Q16

Let

f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx}

. If

f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)

, then

f(4)

is equal to

A

{\log _e}19 - {\log _e}20

B

{\log _e}17 - {\log _e}18

C

{1 \over 2}({\log _e}19 - {\log _e}17)

D

{1 \over 2}({\log _e}17 - {\log _e}19)

Correct Answer

Option D

Solution

$f(x)=\int \dfrac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$

\begin{aligned} & \text{ Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\ & f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\ & =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right)+C \Rightarrow \\\\ & f(3)=\frac{1}{2} \log _{e}\left(\frac{10}{12}\right)+C \\\\ & \because f(3)+\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right) \Rightarrow C=0 \\\\ & f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right) \Rightarrow \\\\ & f(4)=\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right) \end{aligned}

Q17

If

\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C

where C is a constant of integration then :

A A =

{1 \over {54}}

and f(x) = 9(x–1)2

B A =

{1 \over {54}}

and f(x) = 3(x–1)

C A =

{1 \over {81}}

and f(x) = 3(x–1)

D A =

{1 \over {27}}

and f(x) = 9(x–1)2

Correct Answer

Option B

Solution

\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} }

Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)

\Rightarrow \tan \theta = {{x - 1} \over 3}

On Differentiating ...(i) 2(x – 1)dx = 18tan $\theta$ sec2 $\theta$ d $\theta$

\therefore I = \int {{{18\tan \theta {{\sec }^2}\theta d\theta } \over {2 \times 3\tan \theta \times 81{{\sec }^4}\theta }}}

I = {1 \over {27}}\int {{{\cos }^2}\theta \,d\theta } = {1 \over {27}} \times {1 \over 2}\int {\left( {1 + \cos 2\theta } \right)} d\theta

I = {1 \over {54}}\left\{ {\theta + {{\sin 2\theta } \over 2}} \right\} + c

I = {1 \over {54}}\left\{ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {1 \over 2} \times {{2\left( {{{x - 1} \over 3}} \right)} \over {1 + {{\left( {{{x - 1} \over 3}} \right)}^2}}}} \right\} + c

I = {1 \over {54}}\left[ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{3(x - 1)} \over {{x^2} - 2x + 10}}} \right] + c

then

A = {1 \over {54}}

f(x) = 3(x – 1)

Q18

\int {{e^{\sec x}}}

(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx

= esecxf(x) + C then a possible choice of f(x) is :-

A x sec x + tan x + 1/2

B sec x + xtan x - 1/2

C sec x - tan x - 1/2

D sec x + tan x + 1/2

Correct Answer

Option D

Solution

Given

\int {{e^{\sec x}}\left( {\sec x\tan x\,f(x) + (sec\,x\,tan\,x\, + se{c^2}x)} \right)}

= {e^{\sec x}}f(x) + C

Differentiating both sides with respect to x,

{e^{\sec x}}.\sec x\tan x\,f\left( x \right)

+

{e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)

=

{e^{\sec x}}

. f'(x) + f(x)

{e^{\sec x}}

.

\sec x\tan x

$\Rightarrow$

{e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)

=

{e^{\sec x}}

. f'(x) $\Rightarrow$ f'(x) =

{\sec x\tan x + {{\sec }^2}x}

$\therefore$

f(x) = \int {\left( {\left( {\sec x\tan x} \right) + {{\sec }^2}x} \right)dx}

$\therefore$

f(x) = \sec x + \tan x + C

From question we can not find the value of C. So we have to choose any random value of C where (

\sec x + \tan x

) present. Only in option D, (

\sec x + \tan x

) term present. So only possible option is D.

Q19

The integral

\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}}

is equal to (Hence C is a constant of integration)

A -3/4 tan - 4 / 3 x + C

B 3tan–1/3x + C

C –3cot–1/3x+ C

D - 3tan–1/3x + C

Correct Answer

Option D

Solution

\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx

=

\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx}

=

\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx}

Let cot x = t3 $\Rightarrow$ - cosec2x dx = 3t2dt =

- 3\int {{{{t^2}dt} \over {{t^2}}}}

= -3t + C = -3

{{{\cot }^{{1 \over 3}}}x}

+ C = -3

{{{\tan }^{ - {1 \over 3}}}x}

+ C

Q20

If

\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C}

where C is a constant of integration, then the function ƒ(x) is equal to

A

{3 \over {{x^2}}}

B

- {1 \over {6{x^3}}}

C

- {1 \over {2{x^3}}}

D

- {1 \over {2{x^2}}}

Correct Answer

Option C

Solution

I =

\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}}

=

\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}}

Let

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

= t $\Rightarrow$

{{ - 6} \over {{x^7}}}dx = dt

$\Rightarrow$

{{dx} \over {{x^7}}} = - {{dt} \over 6}

$\therefore$ I =

- {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}}

=

- {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C

=

- {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C

=

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

=

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

$\therefore$ f(x) =

{ - {1 \over {2{x^3}}}}