Find $\int\sin 2x\cos 3x\,dx$

$\frac{1}{2} \cos x - \frac{1}{10} \cos 5x + C$

The correct answer is Option (1) → $\frac{1}{2} \cos x - \frac{1}{10} \cos 5x + C$

From the identity $\sin x \cos y = \frac{1}{2} [\sin (x + y) + \sin (x - y)]$

Then $\displaystyle \int \sin 2x \cos 3x \, dx = \frac{1}{2} \left[ \int \sin 5x \, dx - \int \sin x \, dx \right]$

$= \frac{1}{2} \left[ -\frac{1}{5} \cos 5x + \cos x \right] + C$

$= -\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$