Answer
$$\frac{1}{4}$$
Work Step by Step
Given $$ \int_{0}^{\pi / 4} x \sin 2 x d x$$
Let
\begin{align*}
u&=x\ \ \ \ \ \ \ \ dv=\sin 2x\\
du&=dx\ \ \ \ \ \ \ \ v= \frac{-1}{2}\cos 2x
\end{align*}
Then
\begin{align*}
\int_{0}^{\pi / 4} x \sin 2 x d x&= \frac{-x}{2}\cos 2x \bigg|_{0}^{\pi/4}+\frac{1}{2}\int_{0}^{\pi / 4} \cos2 x d x\\
&= \frac{-x}{2}\cos 2x \bigg|_{0}^{\pi/4}+ \frac{1}{4}\sin 2x\bigg|_{0}^{\pi/4}\\
&= \frac{1}{4}
\end{align*}
