#### Need solution for RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 16 textbook solution.

Answer:- $\frac{\alpha \pi}{\sin \alpha}$

Hints:-  You must know the integration rules of trigonometric functions and its limits.

Given:-  $\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x, \quad 0<\alpha<\pi$

Solution : $\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$                                                 ....(1)

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

\begin{aligned} &2 I=\pi \int_{0}^{\pi} \frac{d x}{1+\cos \alpha \sin x} \\ &2 I=\pi \int_{0}^{\pi} \frac{\sec ^{2} \frac{\pi}{2} d x}{\left(1+\tan ^{2} \frac{x}{2}\right)+2 \cos \alpha \tan \frac{x}{2}} \end{aligned}

Put $\tan\frac{x}{2}=t$

\begin{aligned} &\sec ^{2} \frac{\pi}{2} d x=2 d t \\ &2 I=\pi \int_{0}^{\pi} \frac{2 d t}{1+t^{2}+2 t \cos \alpha} \\ &2 I=2 \pi \int_{0}^{\pi} \frac{d t}{(t+\cos \alpha)^{2}+\sin ^{2} \alpha} \end{aligned}

\begin{aligned} &I=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}\left(\frac{t+\cos \alpha}{\sin \alpha}\right)\right]_{0}^{\infty} \\ &=\frac{\pi}{\sin \alpha}\left[\tan ^{-1}(\infty)-\tan ^{-1}(\cot \alpha)\right] \\ &=\frac{\pi}{\sin \alpha}\left[\frac{\pi}{2}-\left(\frac{\pi}{2}-\alpha\right)\right] \\ &=\frac{\alpha \pi}{\sin \alpha} \end{aligned}