QJEEM17Madea3: JEE Main 2017 Mathematics - Class 12 JEEM17Madea3 Solution: The integral equals :

Question

Verified Solution • 2025–26 Session moderate

The integral

\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx

equals :

Accepted Answer

tan x + cot x

=

{{\sin x} \over {\cos x}}

+

{{\cos x} \over {\sin x}}

=

{{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}

=

{1 \over {\sin x\,\,\cos x}}

=

{2 \over {\sin 2x}}

\therefore\,\,\,

\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx

=

\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx

=

\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx

=

\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx

Let sin 2x = t

\Rightarrow

\,\,\,

2 cos2x dx = dt

At x =

{{\pi \over {12}}}

, t = sin

{{\pi \over {6}}}

=

{1 \over 2}

At x =

{{\pi \over 4}}

, t = sin

{{\pi \over 2}}

= 1.

=

{1 \over 2}

\int\limits_{{1 \over 2}}^1 {{t^3}.dt}

=

{{1 \over 2}}

\left[ {{{{t^4}} \over 4}} \right]_{{1 \over 2}}^1

=

{{1 \over 2}}

\times

{{1 \over 4}}

\times

[ 1⁴

-

(

{{1 \over 2}}

)⁴]

=

{{1 \over 8}}

\times

{{15 \over 16}}

=

{{15 \over 128}}

Step-by-Step Verified Solution