#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 5 maths textbook solution.

Answer : $1-\frac{\pi}{4}$

Given : $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}} d x$

Hint : Use the formula of $f(x)$ when $f(x)$ is odd and even

Solution : $I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}} d x \quad-----(1)$

We know that $\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x$

\begin{aligned} &I=\int_{\frac{\pi}{-4}}^{\frac{\pi}{4}} \frac{\tan ^{2}(-x)}{1+e^{-x}} d x \\ &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{-x}} d x------(2) \end{aligned}

\begin{aligned} &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x}{1+e^{x}}+\frac{\tan ^{2} x}{1+e^{-x}} d x \\ &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{\tan ^{2} x+e^{x} \tan ^{2} x}{1+e^{x}} d x \\ &2 I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \tan ^{2} x d x \end{aligned}

We know  $\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$ when $f(x)$ is even

\begin{aligned} &\int_{0}^{\frac{\pi}{4}} \tan ^{2} x d x \\ &\int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x-1\right) d x \quad\left[\sec ^{2} x-1=\tan ^{2} x\right] \end{aligned}

\begin{aligned} &I=(\tan x-x)_{0}^{\frac{\pi}{4}} \\ &I=\left(1-\frac{\pi}{4}\right)-(0-0) \\ &=1-\frac{\pi}{4} \end{aligned}