Show that

Let $I=\int_{\frac{\pi}{2}}^{\pi} \sin ^{7} x d x$      ...(1)

As $\sin ^{7}(-x)=(\sin (-x))^{7}=(-\sin x)^{7}=-\sin ^{7} x$, therefore, $\sin ^{2} x$ is an odd function.

It is known that, if $f(x)$ is an odd function, then $\int_{-a}^{a} f(x) d x=0$

$\therefore I=\int_{\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x=0$