Question

Evaluate $\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$.

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

Answer 1

The correct option is C $\frac{{\pi }^2}{4}$

We are given, $I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx$

let $\cos x=t$, differentiating wrt $x$

we get,

$-\sin xdx=dt$

so, when $x=0\Rightarrow t=\cos 0=1$

& $x=\pi \Rightarrow t=\cos \pi 0=.1$

$I=\frac{\pi }{2}{\int }_1^{-1}\frac{-dt}{1+t^2}$

$I=\frac{\pi }{2}{\int }_{-1}^1\frac{dt}{1+t^2}$

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

$I=\frac{\pi }{2}[{\tan }^{-1}t{]}_{-1}^1$

$=\frac{\pi }{2}[{\tan }^{-1}-{\tan }^{-1}(-1\left)\right]$

$=\frac{\pi }{2}[\frac{\pi }{4}+\frac{\pi }{4}]$

$=\frac{\pi }{2}\times \frac{\pi }{2}$

$I=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx=\frac{{\pi }^2}{4}$