Answer
$$\frac{1}{3} \sin ^{3} x-\frac{2}{5}\sin ^{5} x+\frac{1}{7}\sin ^{7} x+c$$
Work Step by Step
\begin{aligned}
\int \sin ^{2} x \cos ^{5} x d x &=\int \sin ^{2} x \cos ^{4} x \cos x d x \\ &=\int \sin ^{2} x\left(1-\sin ^{2} x\right)^{2} \cos x d x \\
&=\int \sin ^{2} x\left(1-2\sin ^{2} x+\sin ^{4} x\right) \cos x d x \\
&=\int \left(\sin ^{2} x-2\sin ^{4} x+\sin ^{6} x\right) \cos x d x \\
&=\frac{1}{3} \sin ^{3} x-\frac{2}{5}\sin ^{5} x+\frac{1}{7}\sin ^{7} x+c
\end{aligned}
