Answer
$$\int_{0}^{\pi / 2} \int_{0}^{\sin \theta} \theta \ r \ d r \ d \theta=\frac{\pi^{2}}{32}+\frac{1}{8}$$
Work Step by Step
Given$$\int_{0}^{\pi / 2} \int_{0}^{\sin \theta} \theta \ r \ d r \ d \theta$$
so, we have
\begin{aligned}
I=& \int_{0}^{\pi / 2} \int_{0}^{\sin \theta} \theta \ r \ d r \ d \theta \\&=\int_{0}^{\pi / 2}\left[\theta \frac{r^{2}}{2}\right]_{0}^{\sin \theta} d \theta\\\
&=\int_{0}^{\pi / 2} \frac{1}{2} \theta \sin ^{2} \theta \ d \theta \\
\end{aligned}
Since $$ \sin ^{2} \theta=\frac{1}{2}(1-\cos 2 \theta) $$ we get
$$I=\frac{1}{4} \int_{0}^{\pi / 2} \theta (1- \cos 2 \theta) d \theta\\$$
By using integration by partition
let
$$u=\theta, \ \ \ \ \ du=d\theta$$
$$dv=1- \cos 2 \theta, \ \ \ \ \ v=\theta-\frac{1}{2}\sin 2 \theta$$
so, we get
\begin{aligned}
I&=\frac{1}{4} uv |_0^{\pi / 2}-\frac{1}{4} \int_{0}^{\pi / 2} v \ du \\
&=\frac{1}{4}\left[\theta (\theta-\frac{1}{2}\sin 2 \theta)\right]_{0}^{\pi / 2}-\frac{1}{4}\int_{0}^{\pi / 2} (\theta - \frac{1}{2} \sin 2 \theta) d \theta\\
&=\frac{\pi^2}{16}-\frac{1}{4} \left[ \frac{\theta^2}{2} + \frac{1}{4}\cos 2 \theta \right]_{0}^{\pi / 2}\\
&=\frac{\pi^2}{16}-\frac{1}{4} \left[ \frac{\pi^2}{8} + \frac{1}{4}\cos \pi-(0 + \frac{1}{4}\cos 0 ) \right] \\
&=\frac{\pi^2}{16}-\frac{1}{4} \left[ \frac{\pi^2}{8} - \frac{1}{2} \right] \\
&=\frac{\pi^{2}}{32}+\frac{1}{8} \end{aligned}
