Answer
$$-\frac{6}{7}$$
Work Step by Step
Given
$$ \int_{0}^{\pi}\sin 3x\cos 4xdx$$
Use
$$
\sin (a x) \cos (b x)=\frac{1}{2} \sin ((a-b) x)+\frac{1}{2} \sin ((a+b) x) $$
Then
\begin{align*}
\int_{0}^{\pi}\sin 3x\cos 4xdx&=\frac{1}{2}\int_{0}^{\pi}(\sin -x+\sin7x)dx\\
&=\frac{1}{2}\int_{0}^{\pi}( \sin7x-\sin x)dx\\
&=\frac{-1}{2}\left(\frac{1}{7}\cos x-\cos x\right)\bigg|_{0}^{\pi} \\
&=-\frac{6}{7}
\end{align*}
