# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x$

Solution:

$\sin 2 x$ can be written as $\sin (5 x-3 x)$

$\therefore$ The equation now becomes

$\Rightarrow \int \frac{\sin (5 x-3 x)}{\sin 5 x \sin 3 x} d x$

$\sin (A-B)=\sin A \cos B-\cos A \sin B$

$\Rightarrow \int \frac{\sin 5 x \cos 3 x-\cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x$

$\Rightarrow \int \frac{\sin 5 x \cos 3 x}{\sin 5 x \sin 3 x} d x-\int \frac{\cos 5 x \sin 3 x}{\sin 5 x \sin 3 x} d x$

$\Rightarrow \int \frac{\cos 3 x}{\sin 3 x} d x-\int \frac{\cos 5 x}{\sin 5 x} d x$

$\Rightarrow \int \cot 3 x d x-\int \cot 5 x d x$

$\Rightarrow \frac{1}{3} \ln |\sin 3 x|-\frac{1}{5} \ln |\sin 5 x|+c$