Answer
$$-\frac{\pi }{2}+1+\frac{\pi ^2}{8}$$
Work Step by Step
Given $$\int_{0}^{\pi / 2} \int_{0}^{x} x \sin y d y d x$$
Since \begin{align*}
\int_{0}^{\pi / 2} \int_{0}^{x} x \sin y d y d x&=\int_{0}^{\pi / 2} x \left[-\cos y\right]_{0}^{x} d x\\
&=\int_{0}^{\pi / 2}(-x\cos x+x)dx
\end{align*}
integrating by parts
\begin{align*}
u&=x\ \ \ \ \ \ \ \ \ \ dv=-\cos x\\
du&=dx\ \ \ \ \ \ \ \ \ \ v=-\sin xdx
\end{align*}
then
\begin{align*}
\int -x\cos xdx&=-x\sin x+\int \sin xdx\\
&=-x\sin x-\cos x
\end{align*}
Hence
\begin{align*}
\int_{0}^{\pi / 2} \int_{0}^{x} x \sin y d y d x &=\int_{0}^{\pi / 2}(-x\cos x+x)dx\\
&=-x\sin x-\cos x+\frac{1}{2}x^2\bigg|_{0}^{\pi/2}\\
&=-\frac{\pi }{2}+1+\frac{\pi ^2}{8}
\end{align*}
