Answer
$$\iint \limits_{R} x \sec ^{2} y\ d A =2$$
Work Step by Step
Given
$$\iint \limits_{R} x \sec ^{2} y d A, \quad R=\{(x, y) | 0 \leqslant x \leqslant 2,0 \leqslant y \leqslant \pi / 4\}$$
So, we have
\begin{aligned}I&=\int_{0}^{2} \int_{0}^{\pi / 4} x \sec ^{2} y d y d x\\
&=\left[\int_{0}^{2} x \ d x\right]\left[\int_{0}^{\pi / 4} \sec ^{2} y \ d y\right]\\
& =\left[\frac{x^{2}}{2}\right]_{0}^{2} \cdot[\tan y]_{0}^{\pi / 4}\\
&=\left[\frac{2^{2}}{2}-\frac{0^{2}}{2}\right] \cdot\left[\tan \frac{\pi}{4}-\tan 0\right] \\
&=2(1)\\
&=2\end{aligned}
