#### Please solve RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 29 maths textbook solution.

Answer:- $\pi^{2}$

Hints:-  You must know the integral rules of logarithmic functions.

Given:-  $\int_{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x$

Solution : $\int_{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x+\int_{-\pi}^{\pi} \frac{2 x \cdot \sin x}{1+\cos ^{2} x} d x$

$I_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; I_{2}$

$I_{1}=0$        [being an odd function]

\begin{aligned} &I_{2}=2 \int_{0}^{\pi} \frac{2 x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &I_{2}=4 \int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

Where $I_{3}=4 I_{2}+0=4 I$

Let  $I_{3}=\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x$

\begin{aligned} &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin (\pi-x)}{1+\cos ^{2}(\pi-x)} \cdot d x \\ &=\int_{0}^{\pi} \frac{(\pi-x) \cdot \sin x}{1+\cos ^{2} x} \cdot d x \\ &=\int_{0}^{\pi} \frac{\pi \cdot \sin x}{1+\cos ^{2} x} \cdot d x-\int_{0}^{\pi} \frac{x \cdot \sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

\begin{aligned} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x-I_{3} \\ 2 I_{3} &=\pi \int_{0}^{\pi} \frac{\sin x}{1+\cos ^{2} x} \cdot d x \end{aligned}

Put $\cos x = t\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x\rightarrow 0,t=1$

$-\sin x \cdot d x=d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad x \rightarrow \pi, t=-1$

\begin{aligned} &2 I_{3}=-\pi \int_{1}^{-1} \frac{d t}{1+t^{2}} \\ &=\pi\left[\tan ^{-1} x\right]_{-1}^{1} \\ &2 I_{3}=\frac{\pi^{2}}{2} \end{aligned}

\begin{aligned} &I_{3}=\frac{\pi^{2}}{4} \\ &\therefore I=\pi^{2} \end{aligned}