Question

Evaluate : ${\int }_{-\pi }^{\pi }\frac{{\sin }^{2}x}{1+e^x}dx$

Answer 1

We know that

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Hence,

$I={\int }_{-\pi }^{\pi }\frac{{\sin }^{2}x}{1+e^x}dx$

Here $a+b=\pi -\pi =0$

$⟹I={\int }_{-\pi }^{\pi }\frac{{\sin }^2(-x)}{1+e^{-x}}dx$

$⟹I={\int }_{-\pi }^{\pi }\frac{{\sin }^{2}x}{1+e^{-x}}dx$

$⟹I={\int }_{-\pi }^{\pi }\frac{e^x{\sin }^{2}x}{1+e^x}dx$

Hence,

$⟹I+I={\int }_{-\pi }^{\pi }\frac{(1+e^x){\sin }^{2}x}{1+e^x}dx$

$⟹2I={\int }_{-\pi }^{\pi }{\sin }^{2}xdx$

$⟹2I=\frac{1}{2}{\int }_{-\pi }^{\pi }(1-\cos 2x)dx$

$⟹I=\frac{1}{4}{[x-\frac{\sin 2x}{2}]}_{-\pi }^{\pi }$

$⟹I=\frac{\pi }{2}$