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

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

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best_answer

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}

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