$\begin{aligned} \lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)\right. & +\sin ^5\left(\frac{3 \pi}{6 n}\right) \\ & \left.+\ldots+\sin ^5\left(\frac{\pi}{2}\right)\right\}\end{aligned}$
$$
\begin{aligned}
=\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sin ^5\left(\frac{\pi}{6 n}\right)+\sin ^5\left(\frac{2 \pi}{6 n}\right)\right. & +\sin ^5\left(\frac{3 \pi}{6 n}\right) \\
& \left.+\ldots+\sin ^5\left(\frac{3 n \pi}{6 n}\right)\right\} \\
= & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{3 n} \sin ^5\left(\frac{r \pi}{6 n}\right)=\int_0^3 \sin ^5\left(\frac{\pi}{6} x\right) d x
\end{aligned}
$$
Let
$$
\frac{\pi}{6} x=t
$$
For upper limit at $x=3, t=\pi / 2$ and lower limit at $x=0, t=0$ and $d x=\frac{6}{\pi} d t$
So,
$$
\begin{aligned}
\int_0^3 \sin ^5\left(\frac{\pi}{6} x\right) d t & =\frac{6}{\pi} \int_0^{\pi / 2} \sin ^5(t) d t \\
& =\frac{6}{\pi} \times \frac{4 \times 2}{5 \times 3 \times 1}=\frac{16}{5 \pi}
\end{aligned}
$$