Answer
$$\displaystyle{\int_0^{\pi}{\sin 6x\cos 3x}\ dx=\frac{4}{9}}$$
Work Step by Step
$
\displaystyle{I=\int_0^{\pi}{\sin 6x\cos 3x\ dx}\\
I=\int_0^{\pi}{2\sin 3x\cos 3x\cos 3x\,\,dx}\\
I=2\int_0^{\pi}{\sin 3x\cos ^23x\,\,dx}}
$
Consider:
$
\displaystyle\frac{d}{dx}\cos ^3x=-9\cos ^23x\sin 3x
$
$
\displaystyle{I=2\int_0^{\pi}{--\frac{9}{9}\sin 3x\cos ^23x\,\,dx}\\
I=-\frac{2}{9}\int_0^{\pi}{-9\sin 3x\cos ^23x\,\,dx}\\
I=-\frac{2}{9}\int_0^{\pi}{-9\cos ^23x\sin 3x\,\,dx}\\
I=-\frac{2}{9}\left[ \cos ^33x \right] _{0}^{\pi}\\
I=\frac{4}{9}}
$
