The integral _/12^/4 8 2x( x + x)^3,dx equals:

Question

The integral $\displaystyle\int_{\pi/12}^{\pi/4} \dfrac{8\cos 2x}{(	an x + \cot x)^3}\,dx$ 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}}$

Solutions · 1