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Please Solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise Revision Exercise Question 44 Maths Textbook Solution.

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Answer:  \log 2

Hint: To solve this equation we convert tan in form of sec

Given:   \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}|\tan x| d x


\begin{aligned} &I=\int_{-\frac{\pi}{4}}^{0}-\tan x d x+\int_{0}^{\frac{\pi}{4}} \tan x d x \\ & \end{aligned}

I=\int_{0}^{-\frac{\pi}{4}} \tan x d x+\int_{0}^{\frac{\pi}{4}} \tan x d x

\begin{aligned} I &=[\log |\sec x|]_{0}^{-\frac{\pi}{4}}+[\log |\sec x|]_{0}^{\frac{\pi}{4}} \\ & \end{aligned}

=\log \sqrt{2}+\log \sqrt{2} \\

=2 \log \sqrt{2} \\

=2 \log (2)^{\frac{1}{2}} \\

=\log 2


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