Answer
a) $\int \sin ^n(x) d x=-\frac{1}{n} \cos x \sin ^{n-1} x+\frac{n-1}{n} \int \sin ^{n-2} x d x$
b) $\int \sin ^3(x) d x=\frac{2}{3}, \int \sin ^5(x) d x=\frac{8}{15}$
c) $\int_0^{\frac{\pi}{2}} \sin ^{2 n+1}(x) d x=\frac{2}{3} * \ldots \ldots * \frac{2 n-2}{2 n-1} * \frac{2 n}{2 n+1}$
Work Step by Step
$$
\begin{gathered}
\left|\int_0^{\pi / 2} \sin ^n x d x=-\frac{1}{n} \cos (x) \sin ^{n-1}(x)\right|_0^{\pi / 2}+\frac{n-1}{n} \int_0^{\pi / 2} \sin ^{n-2} x d x \\
\therefore \sin ^n \int x d x=\frac{n-1}{n} \int_0^{\frac{\pi}{2}} \sin ^{n-2}(x) d x
\end{gathered}
$$
b) $\int \sin ^3(x) d x=\frac{3-1}{3} \int_0^{\frac{\pi}{2}} \sin (x) d x=\left.\frac{-2}{3} \cos (x)\right|_0 ^{\frac{\pi}{2}}=\frac{-2}{3}[0-1]=\frac{2}{3}$
$$
\int \sin ^5(x) d x=\frac{5-1}{5} \int_0^{\frac{\pi}{2}} \sin ^3 d x=\frac{5-1}{5} * \int \sin ^3(x) d x=\frac{5-1}{5} * \frac{2}{3}=\frac{8}{15}
$$
c) $\int_0^{\frac{\pi}{2}} \sin ^{2 n+1}(x) d x=\frac{2 n+1-1}{n+1} \int_0^{\frac{\pi}{2}} \sin ^{2 n-1}(x) d x=\frac{2 n}{2 n+1} * \frac{2 n-2}{2 n-1} * \sin ^{2 n-3}(x) d x$
finally we will reach to $n=3$
when $\mathrm{n}=3$ then $\gg \int_0^{\frac{\pi}{2}} \sin ^{3-2}(x) d x=\frac{2}{3} * 1=\frac{2}{3}$
$$
\therefore \int_0^{\frac{\pi}{2}} \sin ^{2 n+1}(x) d x=\frac{2}{3} * \ldots \ldots * \frac{2 n-2}{2 n-1} * \frac{2 n}{2 n+1}
$$
