Question

The value of $\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1+\tan x}{1-\tan x}dx$ is ________________.

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1+\tan x}{1-\tan x}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan x}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\tan \left(\frac{\mathrm{\pi }}{4}+x\right)dx \\ ={\left[-\log \left(\cos \left(\frac{\mathrm{\pi }}{4}+x\right)\right)\right]}_0^{\frac{\mathrm{\pi }}{4}} \\ =-\log \left(\cos \left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{4}\right)\right)+\log \left(\cos \left(\frac{\mathrm{\pi }}{4}+0\right)\right) \\ =-\log \left(\cos \left(\frac{2\mathrm{\pi }}{4}\right)\right)+\log \left(\cos \left(\frac{\mathrm{\pi }}{4}\right)\right) \\ =-\log \left(\cos \left(\frac{\mathrm{\pi }}{2}\right)\right)+\log \left(\frac{1}{\sqrt{2}}\right) \\ =-\log \left(0\right)+\log {\left(2\right)}^{-\frac{1}{2}} \\ =-\frac{1}{2}\log \left(2\right) \end{gathered}$$

​Hence, the value of $\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\frac{1+\tan x}{1-\tan x}dx$ is $\underline{-\frac{1}{2}\log \left(2\right)}.$