Answer
$$\int_{\pi / 2}^{\pi} \int_{1}^{2} x \cos x y d y d x=-2$$
Work Step by Step
Given $$\int_{\pi / 2}^{\pi} \int_{1}^{2} x \cos x y d y d x$$
So, we have
\begin{aligned}
I&=\int_{\pi / 2}^{\pi} \int_{1}^{2} x \cos (x y) \ \ d y\ d x\\
&=\int_{\pi / 2}^{\pi} \sin( x y) | _{1}^{2} \ d x\\
&=\int_{\pi / 2}^{\pi} \left( \sin( 2x ) - \sin( x ) \right) \ d x\\
&= \left(-\frac{1}{2} \cos( 2x ) + \cos( x ) \right)_{\pi / 2}^{\pi}\\
&= \left(-\frac{1}{2} \cos( 2\pi ) + \cos( \pi ) \right)- \left(-\frac{1}{2} \cos( \pi ) + \cos(\frac{ \pi}{2} ) \right)\\
&= \left(-\frac{1}{2} -1 \right)- \left(\frac{1}{2} +0 ) \right)\\
&=-\frac{1}{2} -1-\frac{1}{2} \\
&=-2
\end{aligned}
