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

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

Answers (1)

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

Solution:

\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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