Answer
$$\frac{-1}{4}$$
Work Step by Step
\begin{aligned}
\int_{0}^{\pi/4} \sin 3 x \cos 5 x d x &=\int_{0}^{\pi/4} \frac{1}{2}(\sin 8 x+\sin -2 x) d x \\ &=\int_{0}^{\pi/4} \frac{1}{2}(\sin 8 x-\sin 2 x) d x \\ &=\frac{1}{2} \int_{0}^{\pi/4} \sin 8 x d x-\frac{1}{2} \int_{0}^{\pi/4} \sin 2 x d x \\ &=\frac{1}{2} \frac{-\cos 8 x}{8}-\frac{1}{2} \frac{-\cos 2 x}{2}\bigg|_{0}^{\pi/4}\\
&= \frac{-1}{4}
\end{aligned}
