Answer
\begin{align*}
\int \sin ^{2} x \cos ^{3} x d x &= \frac{1}{3}\sin^3 x-\frac{1}{5}\sin^5x+C\\
\int \sin ^{2} x \cos ^{3} x d x &= \frac{1}{30}(7+3 \cos 2 x) \sin ^{3} x+C
\end{align*}
These are equivalent.
Work Step by Step
Since
\begin{align*}
\int \sin ^{2} x \cos ^{3} x d x&=\int \sin ^{2} x \cos ^{2} x\cos x d x\\
&=\int \sin ^{2} x (1-\sin ^{2} x)\cos x d x\\
&=\int ( \sin ^{2} x-\sin ^{4} x)\cos x d x\\
&= \frac{1}{3}\sin^3 x-\frac{1}{5}\sin^5x+C
\end{align*}
and
\begin{align*}
\frac{1}{3}\sin^3 x-\frac{1}{5}\sin^5x&=\left( \frac{1}{3}-\frac{1}{5}\sin^2x \right)\sin^3 x\\
&= \frac{1}{15}\left( 5-3\sin^2x \right)\sin^3 x\\
&= \frac{1}{15}\left( 5-\frac{3}{2}(1-\cos 2x)\right)\sin^3 x\\
&= \frac{1}{30}\left( 10-3(1-\cos 2x)\right)\sin^3 x\\
&=\frac{1}{30}(7+3 \cos 2 x) \sin ^{3} x
\end{align*}
Hence
\begin{align*}
\int \sin ^{2} x \cos ^{3} x d x &= \frac{1}{3}\sin^3 x-\frac{1}{5}\sin^5x+C\\
\int \sin ^{2} x \cos ^{3} x d x &= \frac{1}{30}(7+3 \cos 2 x) \sin ^{3} x+C
\end{align*}
are equivalent to each other.
