1

### JEE Main 2018 (Online) 15th April Evening Slot

If    $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$
(where C is a constant of integration), then the ordered pair (A, B) is equal to :
A
(2,  1)
B
($-$ 2,   $-$1)
C
($-$ 2,  1)
D
(2,   $-$1)

## Explanation

We can write,

7 - 6x - x2 = 16 - (x + 3)2

and ${d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)$

So, $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx$

= $\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$

= $\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$ -
$\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx$

= $\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$ -
$\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx$

= $- 2\sqrt {7 - 6x - {x^2}}$ - ${\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)$ + C

By comparing with the given equation in the question we get,

A = - 2 and B = - 1
2

### JEE Main 2018 (Online) 16th April Morning Slot

If $\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}}$ $\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$ is a constant of integration) then the ordered pair (K, A) is equal to :
A
(2, 1)
B
($-$2, 3)
C
(2, 3)
D
($-$2, 1)

## Explanation

Given,

$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$

Let    tanx = t

$\Rightarrow $$\,\,\, sec2x dx = dt \Rightarrow$$\,\,\,$ dx = ${{dt} \over {{{\sec }^2}x}}$

$\Rightarrow $$\,\,\, dx = {{dt} \over {1 + {{\tan }^2}x}} \Rightarrow$$\,\,\,$ dx = ${{dt} \over {1 + {t^2}}}$

$\therefore\,\,\,$ $\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$

=   $\int {{t \over {1 + t + {t^2}}}} \times {{dt} \over {1 + {t^2}}}$

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

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

=   ${\tan ^{ - 1}}\left( t \right) - \int {{1 \over {1 + t + {t^2} + {1 \over 4} - {1 \over 4}}}} \,\,dt$

=   $x - \int {{1 \over {{t^2} + t + {1 \over 4} + 1 - {1 \over 4}}}} \,\,dt$

=   $x - \int {{1 \over {{{\left( {t + {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}}$

=   $x - {1 \over {{{\sqrt 3 } \over 2}}} + {\tan ^{ - 1}}\left( {{{2t + 1} \over {\sqrt 3 }}} \right) + c$

=   $x - {2 \over {\sqrt 3 }}\,\,{\tan ^{ - 1}}\left( {{{2\tan x + 1} \over {\sqrt 3 }}} \right) + c$

$\therefore\,\,\,$ By comparing

A = 3 and K = 2
3

### JEE Main 2019 (Online) 9th January Morning Slot

For x2 $\ne$ n$\pi$ + 1, n $\in$ N (the set of natural numbers), the integral

$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx}$ is equal to :

(where c is a constant of integration)
A
${\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$
B
${1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c$
C
${1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$
D
${\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$

## Explanation

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

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

$= \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx$

$= \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx$

$= \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx$

put   ${{{x^2} - 1} \over 2} = t$

$\Rightarrow {x^2} - 1 = 2t$

$\Rightarrow 2xdx = 2dt$

$\Rightarrow xdx = dt$

$\therefore$  $\int {\tan t\,dt}$

$= \ln \left| {\sec t} \right| + C$

$= \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$
4

### JEE Main 2019 (Online) 9th January Evening Slot

If   $f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$

$f\left( 0 \right) = 0,$    then the value of $f(1)$ is :
A
$-$ ${1 \over 2}$
B
$-$ ${1 \over 4}$
C
${1 \over 2}$
D
${1 \over 4}$

## Explanation

$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$

$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$

$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$

Let ${x^{ - 5}} + {x^{ - 7}} + 2 = t$

$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$

$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt$

$f(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c$

$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$

$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$

$f\left( 0 \right) = 0$

$\therefore$ $c = 0$

$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$

$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$

NEET