Question

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

Answer 1

$\text{Let}I={\int }_0^{a}f(a-x)dx\cdots \left(1\right)\text{Put}a-x=t\Rightarrow -dx=dt\text{When x= 0, then t=a, when x=a, t= 0}\therefore I={\int }_a^0-f\left(t\right)dt={\int }_0^{a}f\left(t\right)dt[\because {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx]$

Since, Integration is unaffected of change of variable parameter, we have
$I={\int }_0^{a}f\left(t\right)dt={\int }_0^{a}f\left(x\right)dx\cdots \left(2\right)$

From $\left(1\right)\&\left(2\right),\text{we get}$

$\Rightarrow {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx\text{Hence proved.}$

Using the above property, we get

$I_1={\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx={\int }_0^{\pi }\frac{(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)}dx\Rightarrow I_1={\int }_0^{\pi }\frac{(\pi -x)\sin x}{1+{\cos }^{2}x}dx\Rightarrow I_1={\int }_0^{\pi }\frac{\pi \sin x}{1+{\cos }^{2}x}dx-I_1\Rightarrow 2I_1={\int }_0^{\pi }\frac{\pi \sin x}{1+{\cos }^{2}x}dx$

Put $\cos x=z\Rightarrow -\sin xdx=dz$
When $x=0,z=1$ and when $x=\pi ,z=-1$
$\therefore 2I_1={\int }_1^{-1}\frac{-\pi dz}{1+z^2}={\int }_{-1}^1\frac{\pi dz}{1+z^2}=\pi {\left[{\tan }^{-1}z\right]}_{-1}^1=\pi (\frac{\pi }{4}-\frac{-\pi }{4})=\frac{{\pi }^2}{2}\therefore I_1=\frac{{\pi }^2}{4}$