Answer
$$-\frac{\cos ^{5} x}{5}+\frac{\cos ^{7} x}{7}+C $$
Work Step by Step
We integrate as follows:
\begin{align*}
\int \sin ^{3} x \cos ^{4} x d x&=\int \cos ^{4} x\left(1-\cos ^{2} x\right) \sin x d x\\
&=\int \cos ^{4} x \sin x d x-\int \cos ^{6} x \sin x d x\\
&=-\frac{\cos ^{5} x}{5}+\frac{\cos ^{7} x}{7}+C
\end{align*}
Where we used the fact that $\sin^2 x + \cos^2 x=1$
