Answer
$$ \int_{y}^{\pi / 2} \sin ^{3} x \cos y \ d x=\cos^2 y-\frac{1}{3} \cos ^{4}y$$
Work Step by Step
Given $$ \int_{y}^{\pi / 2} \sin ^{3} x \cos y d x $$ So,\begin{aligned} \int_{y}^{\pi / 2} \sin ^{3} x \cos y dx &= \int_{y}^{\pi / 2} \sin ^{2}x \sin x \cos y d x, \ \ \ \small (sin^2 x=1-cos^2x )\\
&=\int_{y}^{\pi / 2}\left(1-\cos ^{2} x\right) \sin x \cos y d x \\
&=\cos y \int_{y}^{\pi / 2}\left( \sin x -\sin x \cos ^{2} x\right) d x \\ &=\cos y \left[ -\cos x+\frac{1}{3} \cos ^{3} x \right]_{y}^{\pi / 2}\\
&=\cos y \left(\cos y-\frac{1}{3} \cos ^{3} y\right) \\
&=\cos^2 y-\frac{1}{3} \cos ^{4}y
\end{aligned}
