Answer
$$-\frac{\cos 5 x}{10}+\frac{\cos x}{2}+C $$
Work Step by Step
Use the formula
$$\int \sin a x \cos b x d x=-\frac{\cos (a+b) x}{2(a+b)}-\frac{\cos (a-b) x}{2(a-b)}+C$$
with $a= 2,\ \ b=3 $ we get
\begin{align*}
\int \sin 2 x \cos 3 x d x&=-\frac{\cos 5 x}{10}-\frac{\cos( -x)}{(-2)}+C \\
&=-\frac{\cos 5 x}{10}+\frac{\cos x}{2}+C
\end{align*}
