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JEE Main 2016 (Online) 9th April Morning Slot

If   $\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,$

where k is a constant of integration, then A + B +C equals :
A
${{21} \over 5}$
B
${{16} \over 5}$
C
${{7} \over 10}$
D
${{27} \over 10}$

Explanation

$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}}$

=  $\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}}$

=  $\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}}$

Let tan x   =   t2

$\Rightarrow $$\,\,\, sec2xdx = 2t dt as sec2x = 1 + tan2x = 1 + t4 = \int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}} = \int {{{{{\sec }^2}x\left( {{{\sec }^2}x\,dx} \right)} \over {2\sqrt {\tan x} }}} = \int {{{\left( {1 + {t^4}} \right)2t\,dt} \over {2t}}} = \int {\left( {1 + {t^4}} \right)} \,dt = t + {{{t^5}} \over 5} + k = \sqrt {\tan x} + {1 \over 5} tan^{{5 \over 2}}x + k By comparing with the given equation, we get A = {1 \over 2}, B = {5 \over 2}, C = {1 \over 5} \therefore\,\,\, A + B + C = {{16} \over 5} 2 MCQ (Single Correct Answer) JEE Main 2016 (Online) 10th April Morning Slot The integral \int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} is equal to : (where C is a constant of integration.) A - 2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C B - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C C - \sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C D 2\sqrt {{{1 + \sqrt x } \over {1 - \sqrt x }}} + C Explanation I = \int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} = \int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} Let 1 + \sqrt x = t \Rightarrow$$\,\,\,$${1 \over {2\sqrt x }}\,dx = dt I = \int {{{2dt} \over {t\sqrt {2t - {t^2}} }}} Again let t = {1 \over z} \Rightarrow$$\,\,\,$ 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$\int {{{ - dz} \over {\sqrt {2z - 1} }}}$

=  $- 2\sqrt {2z - 1} + C$

=   $- 2\sqrt {{2 \over t} - 1} + C$

=  $- 2\sqrt {{{2 - t} \over t}} + C$

=  $- 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$
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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
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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$\log \left| {{{\sin }}{x \over 2}} \right| + C_1$