Answer
\begin{aligned}
\int \cos ^{m} x \sin ^{n} x d x=&-\frac{\cos ^{m+1} x \sin ^{n-1} x}{m+n}+ \frac{n-1}{m+n} \int \cos ^{m} x \sin ^{n-2} x d x
\end{aligned}
Work Step by Step
\begin{aligned}
\int \cos ^{m} x \sin ^{n} x d x&= \int \cos ^{m} x \sin ^{n-1} x\sin x d x
\end{aligned}
Use integration by parts
\begin{aligned}
u&=\cos ^{m} x \sin ^{n-1} x \ \ \ \ \ \ \ \ &dv&= \sin x d x\\
du&=\left[(n-1) \sin ^{n-2} x \cos ^{m+1} x-m \sin ^{n} x \cos ^{m-1} x\right]dx\ \ \ \ \ \ \ \ &v&=- \cos x
\end{aligned}
Then
\begin{aligned}
\int \cos ^{m} x \sin ^{n} x d x&= \int \cos ^{m} x \sin ^{n-1} x\sin x d x\\
&= -\sin ^{n-1} x \cos ^{m+1} x+(n-1) \int \sin ^{n-2} x \cos ^{m+2} x d x-m \int \sin ^{n} x \cos ^{m} x d x, \ \ \ \text{Use} \ \cos^2x+\sin^2x=1\\
&= -\sin ^{n-1} x \cos ^{m+1} x+(n-1) \int \sin ^{n-2} x \cos ^{m} x d x-(n-1) \int \sin ^{n} x \cos ^{m} x d x-m \int \sin ^{n} x \cos ^{m} x d x\\
[1+(n-1)+m] \int \sin ^{n} x \cos ^{m} x d x&=-\sin ^{n-1} x \cos ^{m+1} x+(n-1) \int \sin ^{n-2} x \cos ^{m} x d x\\
[n+m] \int \sin ^{n} x \cos ^{m} x d x&=-\sin ^{n-1} x \cos ^{m+1} x+(n-1) \int \sin ^{n-2} x \cos ^{m} x d x\\
\int \sin ^{n} x \cos ^{m} x d x=-\frac{\sin ^{n-1} x \cos ^{m+1} x}{m+n}+\frac{n-1}{m+n} \int \sin ^{n-2} x \cos ^{m} x d x
\end{aligned}
