#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 46

Answer:   $\frac{\pi}{\sin \alpha}[\pi+\alpha]$

Hint: To solve this equation we convert cos and sin into tan

Given:  $\frac{\pi}{\sin \alpha}[\pi+\alpha]$

Solution:   $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

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

$I =\int_{0}^{\pi} \frac{\pi-x}{1+\cos \alpha \sin x} d x \\$

$I =\int_{0}^{\pi} \frac{\pi}{1+\cos \alpha \sin x} d x-\int_{0}^{\pi} \frac{x}{1+\cos \alpha \sin x} d x$

Let

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

$2 I =\int_{0}^{\pi} \frac{\pi}{1+\cos x \cdot \frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}} d x$

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

$I=\frac{\pi}{2} \int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}-2 \cos \alpha \tan \frac{x}{2}+1} d x$

Let

\begin{aligned} &\tan \frac{x}{2}=t \\ & \end{aligned}

$\Rightarrow>\sec ^{2} \frac{x}{2} d x=2 d t \\$

$x=0 \rightarrow t=\tan 0=0 \\$

$x=x \rightarrow t=\tan \frac{\pi}{2}=\infty$

\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\infty} \frac{2 d t}{t^{2}-2 \cos \alpha t+1} \\ & \end{aligned}

$I=\pi \int_{0}^{\infty} \frac{d t}{(t-\cos \alpha)^{2}+\left(1-\cos ^{2} x\right)} \\$

$I=\pi \int_{0}^{\infty} \frac{d t}{\sin ^{2} \alpha+(t-\cos x)^{2}}$

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

$I=\frac{\pi}{\sin \alpha}\left(\tan ^{-1} \infty-\tan ^{-1}\left(\frac{-\cos \alpha}{\sin \alpha}\right)\right) \\$

$I=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}-\tan ^{-1}(-\cot \alpha)\right)$

\begin{aligned} &=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1}(\cot \alpha)\right) \\ & \end{aligned}

$=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\tan ^{-1} \tan \left(\frac{\pi}{2}+\alpha\right)\right) \\$

$=\frac{\pi}{\sin \alpha}\left(\frac{\pi}{2}+\left(\frac{\pi}{2}+\alpha\right)\right)$

$= \frac{\pi}{\sin \alpha}[\pi+\alpha]$