JEE MAIN - Mathematics (2017 - 8th April Morning Slot - No. 17)
The integral $$\int_{{\pi \over {12}}}^{{\pi \over 4}} {\,\,{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,dx$$ equals :
$${{15} \over {128}}$$
$${{15} \over {64}}$$
$${{13} \over {32}}$$
$${{13} \over {256}}$$
Explanation
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]
= $${{1 \over 8}}$$ $$ \times $$ $${{15 \over 16}}$$ = $${{15 \over 128}}$$
= $${{\sin x} \over {\cos x}}$$ + $${{\cos x} \over {\sin x}}$$
= $${{{{\sin }^2}x + {{\cos }^2}x} \over {\sin x\,\,\cos x}}$$
= $${1 \over {\sin x\,\,\cos x}}$$
= $${2 \over {\sin 2x}}$$
$$\therefore\,\,\,$$ $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {\tan x + \cot x} \right)}^3}}}} \,\,dx$$
= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8\cos 2x} \over {{{\left( {{2 \over {\sin 2x}}} \right)}^3}}}} \,\,dx$$
= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{8{{\sin }^3}\,2x.\cos \,\,2x} \over 8}} \,\,dx$$
= $$\int\limits_{{\pi \over {12}}}^{{\pi \over 4}} {{{\sin }^3}2x.\cos \,2x} \,\,dx$$
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]
= $${{1 \over 8}}$$ $$ \times $$ $${{15 \over 16}}$$ = $${{15 \over 128}}$$
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