Question
Find: $\int \sin 2 x \cos 3 x d x$

Answer

As we know , sin x cos y = $\frac{1}{2}[\sin (x+y)+\sin (x-y)]$
Then , $\int \sin 2 x \cos 3 x d x=\frac{1}{2}\left[\int \sin 5 x d x-\int \sin x d x\right]$
= $\frac{1}{2}\left[-\frac{1}{5} \cos 5 x+\cos x\right]+C$
= $-\frac{1}{10} \cos 5 x+\frac{1}{2} \cos x+C$

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