Question

The integral ${\int }_{\frac{x}{12}}^{\frac{x}{4}}\frac{8\cos 2x}{{(\tan x+\cot x)}^3}dx$ is equal to

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

Answer 1

Answer: (A)

$\frac{15}{128}$

$I=\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{(\tan x+\cot x)}^3}dx$
$=\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})}^3}dx$
$=\underset{\pi /12}{\overset{\pi /4}{\int }}\frac{8\cos 2x}{{\left(\frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x\cos x}\right)}^3}dx$
$=\underset{\pi /12}{\overset{\pi /4}{\int }}8\cos 2x{\left(\sin x\cos x\right)}^{3}dx$
$=\underset{\pi /12}{\overset{\pi /4}{\int }}{\frac{8\cos 2x\left(2\sin x\cos x\right)}{8}}^{3}dx$
$=\underset{\pi /12}{\overset{\pi /4}{\int }}\cos 2x\left({\sin }^{3}2x\right)dx$
$let\sin 2x=t$
$2\cos 2xdx=dt$
$whenx=\pi /12t=1/2$
$x=\pi /4t=1$
$I=\frac{1}{2}\underset{1/2}{\overset{1}{\int }}{t}^{3}dt$
$=\frac{1}{2}{\left(\frac{t^4}{4}\right)}_{1/2}^1$
$=\frac{1}{8}[1-\frac{1}{16}]$
$=\frac{15}{128}$