1

### JEE Main 2016 (Online) 10th April Morning Slot

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

(where C is a constant of integration.)
A
$- 2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C$
B
$- 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$
C
$- \sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$
D
$2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C$

## Explanation

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

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

Let  1 + $\sqrt x$ = t

$\Rightarrow $$\,\,\,$${1 \over {2\sqrt x }}\,dx$ = dt

I  =  $\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}}$

Again let t = ${1 \over z}$

$\Rightarrow $$\,\,\, dt = - {1 \over {{z^2}}}dz \therefore\,\,\, I = 2 \int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}} = 2\int {{{ - dz} \over {\sqrt {2z - 1} }}} = - 2\sqrt {2z - 1} + C = - 2\sqrt {{2 \over t} - 1} + C = - 2\sqrt {{{2 - t} \over t}} + C = - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C 2 MCQ (Single Correct Answer) ### JEE Main 2017 (Offline) Let {I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right). If {I_4} + {I_6} = a{\tan ^5}x + b{x^5} + C, where C is a constant of integration, then the ordered pair \left( {a,b} \right) is equal to A \left( {{1 \over 5},0} \right) B \left( {{1 \over 5}, - 1} \right) C \left( { - {1 \over 5},0} \right) D \left( { - {1 \over 5},1} \right) ## Explanation Given, In = \int {{{\tan }^n}x\,dx,\,\,\,n > 1} \therefore\,\,\, I4 = \int {{{\tan }^4}x\,dx} and I6 = \int {{{\tan }^6}} x\,dx \therefore\,\,\, I = I4 + I6 = \int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx = \int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx = \int {{{\tan }^4}} x.{\sec ^2}x\,dx Let, tanx = t \Rightarrow$$\,\,\,$ sec2x dx = dt

$\therefore\,\,\,$ I = $\int {{t^4}\,dt}$

= ${1 \over 5}$ t5 + C

= ${1 \over 5}$ tan5x + C

$\therefore\,\,\,$ By comparing with the question, we get

A = ${1 \over 5}$,  B = 0
3

### JEE Main 2017 (Online) 8th April Morning Slot

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

## Explanation

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$\log \left| {{{\sin }}{x \over 2}} \right| + C_1$
4

### JEE Main 2017 (Online) 9th April Morning Slot

If $\,\,\,$ f$\left( {{{3x - 4} \over {3x + 4}}} \right)$ = x + 2, x $\ne$ $-$ ${4 \over 3}$, and

$\int {}$f(x) dx = A log$\left| {} \right.$1 $-$ x $\left| {} \right.$ + Bx + C,

then the ordered pair (A, B) is equal to :

(where C is a constant of integration)
A
$\left( {{8 \over 3},{2 \over 3}} \right)$
B
$\left( { - {8 \over 3},{2 \over 3}} \right)$
C
$\left( { - {8 \over 3}, - {2 \over 3}} \right)$
D
$\left( { {8 \over 3}, - {2 \over 3}} \right)$

## Explanation

Given,

f$\left( {{{3x - 4} \over {3x + 4}}} \right)$ = x + 2,    x  $\ne$ $-$ ${4 \over 3}$

Let, ${{3x - 4} \over {3x + 4}}$ = t

$\Rightarrow $$\,\,\, 3x - 4 = 3tx + 4t \Rightarrow$$\,\,\,$ 3x $-$ 3tx = 4t + 4

$\Rightarrow$$\,\,\,$ x = ${{4t + 4} \over {3 - 3t}}$

So, f(t) = ${{4t + 4} \over {3 - 3t}}$ + 2 = ${{10 - 2t} \over {3 - 3t}}$

$\therefore\,\,\,$ f (x) = ${{10 - 2x} \over {3 - 3x}}$

$\therefore\,\,\,$ $\int {f(x)\,dx}$

= $\int {{{2x - 10} \over {3x - 3}}} \,dx$

= $\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} }$

= ${2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}}$

= ${2 \over 3}$ x + ${2 \over 3}$ log $\left| {x - 1} \right|$ $-$ ${{10} \over 3}$ log $\left| {x - 1} \right|$ + C

= ${2 \over 3}$ x $-$ ${8 \over 3}$ log $\left| {x - 1} \right|$ + C

$\therefore\,\,\,$ A = $-$ ${8 \over 3}$ and B = ${2 \over 3}$