Answer
$$\int \sin ^{n} x d x=-\frac{\sin ^{n-1} x \cos x}{n}+\frac{n-1}{n} \int \sin ^{n-2} x d x$$
Work Step by Step
Given
$$\int \sin ^{n} x d x$$
Since
$$\int \sin ^{n-1} x\sin x d x$$
Use integration by parts , let
\begin{aligned}
u&=\sin ^{n-1} x \ \ \ \ \ \ \ \ \ &dv&= \sin x d x\\
du&=(n-1)\sin ^{n-2} x\cos x \ \ \ \ \ \ \ \ \ & v&=- \cos x
\end{aligned}
Then
\begin{aligned}
\int \sin ^{n-1} x\sin x d x &= -\sin^{n-1}x\cos x +(n-1)\int \sin^{n-2}x\cos^2 xdx\\
&= -\sin^{n-1}x\cos x +(n-1)\int \sin^{n-2}x(1-\sin^2 x)dx\\
&= -\sin^{n-1}x\cos x +(n-1)\int \sin^{n-2}xdx+(n-1)\int \sin^n x)dx\\
n\int \sin ^{n-1} x\sin x d x&=-\sin^{n-1}x\cos x +(n-1)\int \sin^{n-2}xdx\\
\int \sin ^{n-1} x\sin x d x&=\frac{-\sin^{n-1}x\cos x}{n} +\frac{(n-1)}{n}\int \sin^{n-2}xdx
\end{aligned}
