Question

${\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(1+tanx\right)dx=$

• A. $\frac{\mathrm{\pi }}{8}{\log }_{e}2$
• B. $\frac{\mathrm{\pi }}{4}{\log }_{2}e$
• C. $\frac{\mathrm{\pi }}{4}{\log }_{e}2$
• D. $\frac{\mathrm{\pi }}{8}{\log }_e\frac{1}{2}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{8}{\log }_{e}2$

Explanation for Correct answer:

Finding the value of the given integral,

Let $I={\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(1+tanx\right)dx\to \left(i\right)$

We know that, ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$

Replace $x\to 0+\frac{\mathrm{\pi }}{4}-x\left[\text{replace by upper limit +lower limit}-x\right]$

$\begin{array}{rcl}I& =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(1+tan\left(0+\frac{\mathrm{\pi }}{4}-x\right)\right)dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(1+\frac{\tan \left(\frac{\mathrm{\pi }}{4}\right)-\tan x}{1+\tan \left(\frac{\mathrm{\pi }}{4}\right)\tan x}\right)dx\left[\because \tan \left(A-B\right)=\frac{\tan A-\tan B}{1-\tan A\tan B}\right]\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(1+\frac{1-\tan x}{1+\tan x}\right)dx\left[\because t\mathrm{an}\left(\frac{\mathrm{\pi }}{4}\right)=1\right]\\ & & \\ & =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(\frac{1+\tan x+1-\tan x}{1+\tan x}\right)dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log \left(\frac{2}{1+\tan x}\right)dx\\ & =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log 2-\log (1+\tan x)dx\to \left(ii\right)\end{array}$

Adding equations $\left(i\right)$ and $\left(ii\right)$, we get

$\begin{array}{rcl}2I& =& {\int }_0^{\frac{\mathrm{\pi }}{4}}\log (1+\tan x)dx+{\int }_0^{\frac{\mathrm{\pi }}{4}}\log 2-\log (1+\tan x)dx\\ & =& \log 2{\int }_0^{\frac{\mathrm{\pi }}{4}}dx\\ & =& \log 2{\left[x\right]}_0^{\frac{\mathrm{\pi }}{4}}\\ & =& \frac{\mathrm{\pi }}{4}\log 2\end{array}$

$$\begin{gathered} 2I=\frac{\mathrm{\pi }}{4}\log 2 \\ I=\frac{\mathrm{\pi }}{8}\mathrm{lo}{g}_{e}2 \end{gathered}$$

Hence, option (A) is the correct answer.