#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 49

Answer:  $-\frac{\pi}{2}$

Hint: To solve this equation we use By Part

Given:  $\int_{0}^{\pi} \cos 2 x \log \sin x d x$

Solution:  $\int f(x) g(x) d x\\$

$\downarrow$

By Parts

\begin{aligned} &f(x) g(x) d x=\int \frac{d(f x)}{d x} \int g(x) d x(d x) \\ & \end{aligned}

$f(x)=\log \sin x \\$

$g(x)=\cos 2 x$

\begin{aligned} &\Rightarrow\left[\log (\sin x) \int \cos 2 x d x-\int \frac{d(\log \sin x)}{d x} \int \cos 2 x d x d x\right]_{0}^{\pi} \\ & \end{aligned}

${\left[\log (\sin x) \frac{\sin 2 x}{2}-\int \frac{1}{\sin x} \cos x \cdot \frac{\sin 2 x}{x} d x\right]_{0}^{\pi}}$

\begin{aligned} &\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int 2 \cos ^{2} x d x\right]_{2}^{\pi} \\ & \end{aligned}

$\Rightarrow\left[\frac{1}{2} \sin 2 x \cdot \log \sin x-\frac{1}{2} \int(\cos 2 x-1)\right]_{2}^{\pi} \\$

$\Rightarrow\left[\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2}\left(\frac{\sin 2 x}{2}+x\right)\right]_{0}^{\pi}\right]$

\begin{aligned} &I=\left[\frac{1}{2} \sin 2 x \cdot \log (\sin x)-\frac{1}{2} \frac{\sin 2 x}{2}-\frac{x}{2}\right]_{0}^{\pi} \\ & \end{aligned}

$\frac{1}{2} \sin 2 \pi \cdot \log (\sin \pi)-\frac{1}{2} \frac{\sin 2 \pi}{2}-\frac{\pi}{2}-\frac{1}{2} \sin 0 \log \sin 0+\frac{1}{2} \sin 0+0 \\$

$=0-0-\frac{\pi}{2}-0+0+0 \\$

$=-\frac{\pi}{2}$