Question

Integrate ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\mathrm{x}\right)\right)dx$.

Answer 1

Solve the given integral by using properties

Using the Property of definite Integral ${\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^{\mathrm{a}}\mathrm{f}\left(\mathrm{a}-\mathrm{x}\right)\mathrm{dx}$

Let, $\mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\mathrm{x}\right)\right)$

By using the property, we get,

$$\begin{gathered} \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)\right)\left[\because \tan \left(\mathrm{a}-\mathrm{b}\right)=\frac{\tan \left(\mathrm{a}\right)-\tan \left(\mathrm{b}\right)}{1+\tan \left(\mathrm{a}\right)\tan \left(\mathrm{b}\right)}\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\frac{1-\tan \left(\mathrm{x}\right)}{1+\tan \left(\mathrm{x}\right)}\right) \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\frac{2}{1+\tan \left(\mathrm{x}\right)}\right)\left[\because \log \left(\frac{\mathrm{a}}{\mathrm{b}}\right)=\log \left(\mathrm{a}\right)-\log \left(\mathrm{b}\right)\right] \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(2\right)-{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\mathrm{x}\right)\right) \\ \Rightarrow \mathrm{I}={\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(2\right)-\mathrm{I} \\ \Rightarrow 2\mathrm{I}=\frac{\mathrm{\pi }}{4}\log \left(2\right) \\ \Rightarrow \mathrm{I}=\frac{\mathrm{\pi }}{8}\log \left(2\right) \end{gathered}$$

Hence, integration of ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(1+\tan \left(\mathrm{x}\right)\right)$ is $\frac{\mathrm{\pi }}{8}\log \left(2\right)$.