Answer
\begin{aligned}
\int_{0}^{\pi / 2} \sin ^{7} y d y =\frac{16}{35}
\end{aligned}
Work Step by Step
Given $$ \int_{0}^{\pi / 2} \sin ^{7} y d y $$
So, we have
\begin{aligned}
I&= \int_{0}^{\pi / 2} \sin ^{7} y d y\\
&=\int_{0}^{\pi / 2} \sin ^{6} y \sin y d y\\
&=\int_{0}^{\pi / 2}\left(1-\cos ^{2} y\right)^{3} \sin y d y\\
&=\int_{0}^{\pi / 2}\left(1-3\cos ^{2}y +3\cos ^{4}y -\cos ^{6} y\right) \sin y d y\\
&=\int_{0}^{\pi / 2} \sin y d y-3 \int_{0}^{\pi / 2} \cos ^{2} y \sin y d y\\&+3 \int_{0}^{\pi / 2} \cos ^{4} y \sin y d y-\int_{0}^{\pi / 2} \cos ^{6} y \sin y d y\\
&=\left[-\cos y+3 \frac{\cos ^{2} y}{3}-3 \frac{\cos ^{5} y}{5}+\frac{\cos ^{2} y}{7}\right]_{0}^{\pi / 2}\\
&=\left[-\cos \pi / 2+3 \frac{\cos ^{2} \pi / 2}{3}-3 \frac{\cos ^{5} \pi / 2}{5}+\frac{\cos ^{2} \pi / 2}{7}\right] -\left[-\cos 0+3 \frac{\cos ^{2} 0}{3}-3 \frac{\cos ^{5} 0}{5}+\frac{\cos ^{2} 0}{7}\right] \\
&=(0)-\left(-1+1-\frac{3}{5}+\frac{1}{7}\right)\\
&=\frac{16}{35}
\end{aligned}
