Answer
$$\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r d r d \theta=\frac{\pi}{2}$$
Work Step by Step
Given$$\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r d r d \theta$$
So, we have
\begin{align}
\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r d r d \theta& =\int_{0}^{\pi / 2}\left[\frac{r^{2}}{2}\right]_{0}^{2 \cos \theta} d \theta\\
&=\int_{0}^{\pi / 2} 2 \cos ^{2} \theta d \theta
\end{align}
Since $$\cos ^{2}\theta =\frac{1}{2}(1+\cos 2\theta)$$
So, we get
\begin{align}
\int_{0}^{\pi / 2} \int_{0}^{2 \cos \theta} r d r d \theta &
=\int_{0}^{\pi / 2} (1+\cos 2\theta) d \theta\\
&=\left[\theta-\frac{1}{2} \sin 2 \theta\right]_{0}^{\pi / 2}\\
&=\frac{\pi}{2}
\end{align}
