1

### JEE Main 2019 (Online) 10th January Morning Slot

Let n $\ge$ 2 be a natural number and $0 < \theta < {\pi \over 2}.$ Then $\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$ is equal to - (where C is a constant of integration)
A
${n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$
B
${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$
C
${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$
D
${n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$

## Explanation

$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$

$= \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$

Put $1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t$

So ${{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt$

Now  ${1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt}$

$= {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C$

$= {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C$
2

### JEE Main 2019 (Online) 10th January Evening Slot

If  $\int \,$x5.e$-$4x3 dx = ${1 \over {48}}$e$-$4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
A
$-$2x3 $-$ 1
B
$-$ 2x3 + 1
C
4x3 + 1
D
$-$4x3 $-$ 1

## Explanation

$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$

Put  ${x^3} = t$

$3{x^2}\,dx = dt$

$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$

${1 \over 3}\int {t.{e^{ - 4t}}dt}$

${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]$

$- {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c$

${{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c$

$\therefore$  $f(x) = - 1 - 4{x^3}$
3

### JEE Main 2019 (Online) 11th January Evening Slot

If   $\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$ = f(x) $\sqrt {2x - 1}$ + C, where C is a constant of integration, then f(x) is equal to :
A
${2 \over 3}$ (x $-$ 4)
B
${1 \over 3}$ (x + 4)
C
${1 \over 3}$ (x + 1)
D
${2 \over 3}$ (x + 2)

## Explanation

$\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt$

$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } }$

$= {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c$

$= \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c$

$\Rightarrow f\left( x \right) = {{x + 4} \over 3}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

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

## Explanation

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

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

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

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