Answer
$$\int_{\pi / 6}^{\pi / 3} \sin 6 x \cos 4 x d x =\frac{3}{10}$$
Work Step by Step
$$
\int_{\pi / 6}^{\pi / 3} \sin 6 x \cos 4 x d x
$$
Since $$
\sin m x \cos n x=\frac{1}{2}(\sin [(m-n) x]+\sin [(m+n) x])
$$
Then
\begin{align*}
\int_{\pi / 6}^{\pi / 3} \sin 6 x \cos 4 x d x&=\frac{1}{2}\int_{\pi / 6}^{\pi / 3} (\sin 2 x +\sin 10 x )d x\\
&=\frac{1}{2} \left(\frac{-1}{2}\cos 2 x -\frac{1}{10}\cos 10 x \right)\bigg|_{\pi / 6}^{\pi / 3}\\
&=\frac{3}{10}
\end{align*}
