Question

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

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

Answer 1

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

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

$I={\int }_{\pi /12}^{\pi /4}\frac{2.8\cos 2x{\sin }^{3}2x}{2}\frac{dx}{8}$

Put $\sin 2x=t\Rightarrow 2\cos 2xdx=dt$

When $x=\frac{\pi }{12},t=\frac{1}{2}$

$x=\frac{\pi }{4},t=1$

$I={\int }_{\frac{1}{2}}^1\frac{t^3}{2}dt$

$=\frac{1}{2}{\left(\frac{t^4}{4}\right)}_{\frac{1}{2}}^1$

$=\frac{1}{8}(1-\frac{1}{16})$

$=\frac{15}{128}$.