Question

Evaluate the definite integral ${\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$

Answer 1

Let $I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$ .............. (1)
$I={\int }_0^{\pi }\left\{\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}\right\}dx,(\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right)$
$\Rightarrow I={\int }_0^{\pi }\left\{\frac{-(\pi -x)\tan x}{-(\sec x+\tan x)}\right\}dx$
$\Rightarrow I={\int }_0^{\pi }\frac{(\pi -x)\tan x}{\sec x+\tan x}dx$ ............. (2)
Adding (1) and (2), we obtain
$2I={\int }_0^{\pi }\frac{\pi \tan x}{\sec x+\tan x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{\sin x+1-1}{1+\sin x}dx$
$\Rightarrow 2I=\pi {\int }_0^{\pi }1\cdot dx-\pi {\int }_0^{\pi }\frac{1}{1+\sin x}dx$
$\Rightarrow 2I=\pi [x{]}_0^{\pi }-\pi {\int }_0^{\pi }\frac{1-\sin x}{{\cos }^{2}x}dx$
$\Rightarrow 2I={\pi }^2-\pi {\int }_0^{\pi }({\sec }^{2}x-\tan x\sec x)dx$
$\Rightarrow 2I={\pi }^2-\pi [\tan x-\sec x{]}_0^{\pi }$
$\Rightarrow 2I={\pi }^2-\pi [\tan \pi -\sec \pi -\tan 0+\sec 0]$
$\Rightarrow 2I={\pi }^2-\pi [0-(-1)-0+1]$
$\Rightarrow 2I={\pi }^2-2\pi$
$\Rightarrow 2I=\pi (\pi -2)$
$\Rightarrow I=\frac{\pi }{2}(\pi -2)$