Question

Evaluate: ${\int }_0^{\pi }\frac{4x\sin x}{1+{\cos }^{2}x}dx.$

Answer 1

Let $I={\int }_0^{\pi }\frac{4(\pi -x)sin\left(\pi \right)-x}{1+co{s}^2(\pi -x)}dx$

We know that $\int f(a-x)=\int f\left(a\right)-\int f\left(x\right)$

$I={\int }_0^{\pi }\frac{4\pi sinx}{1+co{s}^{2}x}-{\int }_0^{\pi }\frac{4xsinx}{1+co{s}^{2}x}dx$

or, $I={\int }_0^{\pi }\frac{4\pi sinx}{1+co{s}^{2}x}-I$

$2I=4\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx$

$2I=4\pi .2\times {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx$ $\left\{Applying{\int }_0^{2a}f\right(x)dx=2{\int }_0^{a}f(x\left)dxiff\right(2a-x)=f(x\left)\right\}$

$I=4\pi {\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx$

Put $cosx=t,$ $\Rightarrow -sinxdx=dt$

As well for $x=0,$ $x=\frac{\pi }{2}$

$t=1,t=0$

Therefore, $I=4\pi {\int }_1^0\frac{-dt}{1+t^2}$

$I=4\pi {\int }_0^1\frac{dt}{1+t^2}$

$\left\{\begin{array}{c}{\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\end{array}\right\}$

$I=4\pi [ta{n}^{-1}t{]}_0^1$

$=4\pi [ta{n}^{-1}1-ta{n}^{-1}0]$

$=4\pi \times \frac{\pi }{4}={\pi }^2.$