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

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Solution

Let $I = \int \frac{sin 2 x}{sin 5 x . sin 3 x} d x$

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

$= \int (\frac{sin 5 x . cos 3 x - cos 5 x . sin 3 x}{sin 5 x . sin 3 x}) d x$
$= \int (cot 3 x - cot 5 x) d x$
$= \frac{1}{3} log | sin 3 x | - \frac{1}{5} log | sin 5 x | + c$

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