# Get Answers to all your Questions

#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 13 maths textbook solution

Answer: $\frac{-1}{2} \log _{e} 2$

Hint: Use $\tan \left(\frac{\pi}{4}+x\right)$

Given: $\int_{0}^{\pi / 4} \frac{1+\tan x}{1-\tan x}{\mathrm{dx}}$

Solution:

$\mathrm{I}=\int_{0}^{\pi / 4} \frac{1+\tan x}{1-\tan x} \mathrm{dx}$

$=\int_{0}^{\pi / 4} \frac{\tan \left(\frac{\pi}{4}\right)+\tan x}{1-\tan \left(\frac{\pi}{4}\right) \tan x} d x$                            $\left[\because \tan \left(\frac{\pi}{4}\right)=1\right]$

\begin{aligned} &=\int_{0}^{\pi / 4} \tan \left(\frac{\pi}{4}+x\right) d x \\ \\&=\int_{0}^{\pi / 4} \tan \left(\frac{\pi}{4}+x\right) d x \end{aligned}                                    $\left[\because \tan \left(\frac{\pi}{4}+x\right)=\frac{\tan \pi / 4+\tan x}{1-(\tan \pi / 4)(\tan x)}\right]$

\begin{aligned} &=\log \left[\sec \left(\frac{\pi}{4}+x\right)\right]_{0}^{\frac{\pi}{4}} \\\\ &=\left[\log \sec \left(\frac{\pi}{2}\right)-\log \sec \left(\frac{\pi}{4}\right)\right] \end{aligned}

\begin{aligned} & \\\\ &=0-\frac{1}{2} \log _{e}(2) \\\\ &=\frac{-1}{2} \log _{e} 2 \end{aligned}