Question

The integral ${\int }_{\pi /12}^{\pi /4}\frac{8\cos 2x}{{(\tan x+\cot x)}^3}dx$ equals:

• A. $\frac{15}{128}$
• B. $\frac{15}{64}$
• C. $\frac{13}{32}$
• D. $\frac{13}{256}$

Answer 1

The correct option is A $\frac{15}{128}$

${\int }_{\pi /12}^{\pi /4}\frac{8\cos 2x}{{(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})}^3}dx$

$\Rightarrow {\int }_{\pi /12}^{\pi /4}\frac{8\cos 2x}{{({\sin }^{2}x+{\cos }^{2}x)}^3}{\left(\sin x\cos x\right)}^{3}dx$

$\Rightarrow {\int }_{\pi /12}^{\pi /4}8\cos 2x{\left(2\sin x\cos x\right)}^3\times \frac{1}{2^3}dx$

$\Rightarrow {\int }_{\pi /12}^{\pi /4}\cos 2x{\sin }^{3}2xdx$

$\sin 2x=t$

$2\cos 2xdx=dt$

$\Rightarrow {\int }_{1/2}^1\frac{t^3}{2}dt$

$\Rightarrow \frac{1}{2}\frac{{\left[t^4\right]}_{1/2}^1}{4}$

$\Rightarrow \frac{1}{8}[1-\frac{1}{16}]$

$=\frac{1}{8}\times \frac{15}{16}=\frac{15}{128}$