Prove that ${\int }_0^{\mathrm{a}}\text{⨍ (x)dx=}{\int }_0^{\mathrm{a}}⨍ (\mathrm{a}-\mathrm{x})\mathrm{dx}$, hence evaluate ${\int }_0^{\mathrm{\pi }}\frac{\mathrm{x} \sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}$.

From given information, 

Let

$$\begin{gathered} \mathrm{t}=\mathrm{a}-\mathrm{x} \\ \Rightarrow \mathrm{dt}=-\mathrm{dx} \end{gathered}$$

When $\mathrm{x}=0, \mathrm{t}=\mathrm{a}$

when $\mathrm{x}=\mathrm{a}, \mathrm{t}=0$

Putting the value of x in LHS

$$\begin{gathered} {\int }_{\mathrm{a}}^0 f(\mathrm{a}-\mathrm{t}) (-\mathrm{dt}) \\ =-{\int }_{\mathrm{a}}^0 f(\mathrm{a}-\mathrm{t}) \left(\mathrm{dt}\right) \\ ={\int }_0^{\mathrm{a}} f(\mathrm{a}-\mathrm{t}) \left(\mathrm{dt}\right) \\ ={\int }_0^{\mathrm{a}} f(\mathrm{a}-\mathrm{x}) \left(\mathrm{dx}\right) \left(\because {\int }_{\mathrm{a}}^{\mathrm{b}}\mathit{ }\mathrm{f}\left(\mathrm{t}\right) \mathrm{dt}={\int }_{\mathrm{a}}^{\mathrm{b}} f\left(\mathrm{x}\right) \mathrm{dx}\right) \\ =\mathrm{RHS} \end{gathered}$$

Using this we can solve the given question as follows:

$$\begin{gathered} \mathrm{I}=\underset{0}{\overset{\mathrm{\pi }}{\int }} f\left(\mathrm{x}\right)\mathrm{dx}=\underset{0}{\overset{\pi }{\int }} f(\mathrm{\pi }-\mathrm{x}) \mathrm{dx} \\ \Rightarrow 2\mathrm{I}=\underset{0}{\overset{\pi }{\int }} f\left(\mathrm{x}\right) \mathrm{dx}+\underset{0}{\overset{\mathrm{\pi }}{\int }} f(\mathrm{\pi }-\mathrm{x}) \mathrm{dx}=\underset{0}{\overset{\mathrm{\pi }}{\int }} \frac{\mathrm{x} \sin \mathrm{x}}{1+{\cos }^2 \mathrm{x}}\mathrm{dx}+\underset{0}{\overset{\mathrm{\pi }}{\int }} \frac{(\mathrm{\pi }-\mathrm{x})\sin (\mathrm{\pi }-\mathrm{x})}{1+{\cos }^2(\mathrm{\pi }-\mathrm{x})}\mathrm{dx} \\ \Rightarrow 2\mathrm{I}=\underset{0}{\overset{\pi }{\int }}\frac{x \sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}+\underset{0}{\overset{\pi }{\int }}\frac{(\mathrm{\pi }-\mathrm{x})\sin \mathrm{x}}{1+{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx} \\ \Rightarrow 2\mathrm{I}=\underset{0}{\overset{\pi }{\int }} \frac{\mathrm{\pi } \sin \mathrm{x}}{1+{\cos }^2 \mathrm{x}}\mathrm{dx} \end{gathered}$$

Let, $\cos \mathrm{x}=\mathrm{t} \Rightarrow -\sin \mathrm{x} \mathrm{dx}=\mathrm{t}$

$$\begin{gathered} \Rightarrow 2\mathrm{I}=-\underset{1}{\overset{-1}{\int }}\frac{\mathrm{\pi }}{1+{\mathrm{t}}^2}\mathrm{dt}=-\mathrm{\pi }{\left[{\tan }^{-1}\mathrm{t}\right]}_1^{-1}=-\mathrm{\pi }\left(-\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\pi }}{4}\right)=\frac{{\mathrm{\pi }}^2}{2} \\ \therefore \mathrm{I}=\frac{{\mathrm{\pi }}^2}{4} \end{gathered}$$

Therefore, ${\int }_0^{\mathrm{a}}\text{⨍ }\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\text{⨍ }(\mathrm{a}-\mathrm{x})\mathrm{dx}$ has proven and from that we have evaluated ${\int }_0^{\mathrm{\pi }}\frac{\mathrm{x} \sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\mathrm{dx}=\frac{{\mathrm{\pi }}^2}{4}.$