Question

$\underset{0}{\overset{\mathrm{\pi }}{\int }}\frac{x\tan x}{\sec x+\tan x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_0^{\mathrm{\pi }}\frac{x\tan x}{secx+\tan x}dx...\left(\mathrm{i}\right) \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\tan \left(\mathrm{\pi }-x\right)}{sec\left(\mathrm{\pi }-x\right)+\tan \left(\mathrm{\pi }-x\right)}dx \\ ={\int }_0^{\mathrm{\pi }}\frac{\left(\mathrm{\pi }-x\right)\tan x}{secx+\tan x}dx...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 2\mathrm{I}={\int }_0^{\mathrm{\pi }}\frac{\mathrm{\pi }\tan \mathrm{x}}{\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{x}+\tan \mathrm{x}}d\mathrm{x} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{\mathrm{sinx}}{1+\mathrm{sinx}}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1+\mathrm{sinx}-1}{1+\mathrm{sinx}}\mathrm{dx} \\ =\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left[1-\frac{1}{1+\mathrm{sinx}}\right]\mathrm{dx} \\ =\mathrm{\pi }{\left[\mathrm{x}\right]}_0^{\mathrm{\pi }}-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{1}{1+{\displaystyle \frac{2\tan {\displaystyle \frac{\mathrm{x}}{2}}}{1+{\tan }^2{\displaystyle \frac{\mathrm{x}}{2}}}}}\mathrm{dx} \\ ={\mathrm{\pi }}^2-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{{\sec }^2{\displaystyle \frac{\mathrm{x}}{2}}}{1+{\tan }^2{\displaystyle \frac{\mathrm{x}}{2}}+2\tan {\displaystyle \frac{\mathrm{x}}{2}}}\mathrm{dx} \\ ={\mathrm{\pi }}^2-\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\frac{{\sec }^2{\displaystyle \frac{\mathrm{x}}{2}}}{{\left(1+\tan \frac{\mathrm{x}}{2}\right)}^2}\mathrm{dx} \\ ={\mathrm{\pi }}^2+\mathrm{\pi }{\left[\frac{2}{1+\tan {\displaystyle \frac{\mathrm{x}}{2}}}\right]}_0^{\mathrm{\pi }} \\ ={\mathrm{\pi }}^2+\mathrm{\pi }\left(0-2\right) \\ ={\mathrm{\pi }}^2-2\mathrm{\pi } \\ =\mathrm{\pi }\left(\mathrm{\pi }-2\right) \\ \mathrm{Hence}\mathit{}I=\frac{\mathrm{\pi }}{2}\left(\mathrm{\pi }-2\right) \end{gathered}$$