Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin 2x\cdot {\sin }^{2}x}{1+{\sin }^{8}x}dx$ is equal to

• A. $1$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{8}$
  

Answer 1

The correct option is D $\frac{\pi }{8}$


$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{2\sin x\cos x\cdot {\sin }^{2}x}{1+{\sin }^{8}x}dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{2{\sin }^{3}x\cos x}{1+{\sin }^{8}x}dx$
Let ${\sin }^{4}x=t\Rightarrow 4{\sin }^{3}x\cos x=dt$
$=\underset{0}{\overset{1}{\int }}\frac{1}{2}\frac{dt}{1+t^2}=\frac{1}{2}{\tan }^{-1}t{|}_0^1=\frac{1}{2}{\tan }^{-1}1-0=\frac{\pi }{8}$