Question

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

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

Answer 1

Answer: (A)

$\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}$$\frac{cos2x}{{\left(\frac{1}{sin2x}\right)}^3}=$$\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}$$cos2x\times sin2x.si{n}^2\left(2x\right)dx$
$=\frac{1}{4}\int sin4x.\left(1-cos4x\right)dx$
$=\frac{1}{4}$$\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}$ sin 4x - $\underset{\frac{\pi }{12}}{\overset{\frac{\pi }{4}}{\int }}$ sin 8x