Answer
$$
\int x \sin ^{2} x \cos x d x =\frac{1}{3} x \sin ^{3} x+\frac{1}{3} \cos x-\frac{1}{9} \cos ^{3} x+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int x \sin ^{2} x \cos x d x
$$
Let
$$
\left[\begin{array}{c}
u=x, \quad d v=\sin ^{2} x \cos x d x \\
d u=d x \quad , v=\frac{1}{3} \sin ^{3} x
\end{array}\right]
$$
Integrating by parts, we get
$$
\begin{aligned}
\int x \sin ^{2} x \cos x d x &=\frac{1}{3} x \sin ^{3} x-\int \frac{1}{3} \sin ^{3} x d x \\
&=\frac{1}{3} x \sin ^{3} x-\frac{1}{3} \int\left(1-\cos ^{2} x\right) \sin x d x \\
& \quad\quad\quad\quad \text{let}\left[\begin{array}{c}
u=\cos x, \\
d u=-\sin x d x
\end{array}\right] \text{then} \\
&=\frac{1}{3} x \sin ^{3} x+\frac{1}{3} \int\left(1-y^{2}\right) d y \\
&=\frac{1}{3} x \sin ^{3} x+\frac{1}{3} y-\frac{1}{9} y^{3}+C \\
&=\frac{1}{3} x \sin ^{3} x+\frac{1}{3} \cos x-\frac{1}{9} \cos ^{3} x+C
\end{aligned}
$$
where $C$ is an arbitrary constant.
