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Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 7 textbook solution.

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Answer : \frac{\pi}{3}

Given : \int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}} d x

Hint : You must know about the definite integrals

Solution : I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}} d x \quad-----(1)

\begin{aligned} &I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{-\tan x}} d x------(2) \\ &\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x \end{aligned}

Add (1) and (2)

\begin{aligned} &2 I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}}+\frac{e^{\tan x}}{1+e^{\tan x}} d x \\ &2 I=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}} 1 \cdot d x \\ &2 I=\frac{2 \pi}{3} \\ &I=\frac{\pi}{3} \end{aligned}

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