Answer
$0$
Work Step by Step
Given: $\int_{-1}^{1}\int_{0}^{\pi/2} x \sin (\sqrt y) dy dx$
$\int_{-1}^{1}\int_{0}^{\pi/2} x \sin (\sqrt y) dy dx=\int_{-1}^{1} x dx \int_{0}^{\pi/2} \sin (\sqrt y) dx$
Then, we have
$\int_{-1}^{1} x dx \int_{0}^{\pi/2} \sin (\sqrt y) dx=\int_{-1}^{1} x dx$
Hence, $[\dfrac{x^2}{2}|_{-1}^1=|\dfrac{1^2}{2}-\dfrac{(-1)^2}{2}]=0$
