Integral exercise depend on $\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$

Exercise:

Compute, $\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}$

Solution: Let $I=\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}$

So $I=\int_{0}^{\pi }{\frac{\left( \pi -x \right)\sin \left( \pi -x \right)}{1+{{\cos }^{2}}\left( \pi -x \right)}dx}=\int_{0}^{\pi }{\frac{\left( \pi -x \right)\sin x}{1+{{\left( -\cos x \right)}^{2}}}dx}=\int_{0}^{\pi }{\frac{\pi \sin x-x\sin x}{1+{{\cos }^{2}}x}dx}$

$\Rightarrow \int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}=\int_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx-\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}}$

$\Rightarrow 2\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx}=\int_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx}$

Let $u=\cos x\Leftrightarrow du=-\sin xdx$

So $$\int_{0}^{\pi }{\frac{\pi \sin x}{1+{{\cos }^{2}}x}dx}=-\pi \int_{u\left( 0 \right)}^{u\left( \pi  \right)}{\frac{du}{1+{{u}^{2}}}}=-\pi \int_{1}^{-1}{\frac{du}{1+{{u}^{2}}}=\pi \int_{-1}^{1}{\frac{du}{1+{{u}^{2}}}}}$$

$=\pi \arctan \left( u \right)_{-1}^{1}=\pi \left( \arctan \left( 1 \right)-\arctan \left( -1 \right) \right)=\pi \left( \frac{\pi }{4}+\frac{\pi }{4} \right)=\pi \left( \frac{\pi }{2} \right)=\frac{{{\pi }^{2}}}{2}$

Hence $\int_{0}^{\pi }{\frac{x\sin x}{1+{{\cos }^{2}}x}dx=\frac{{{\pi }^{2}}}{4}}$