Answer
$$ \frac{1}{2}\left(x^{2} \sin x^{2}+\cos x^{2}\right)+C$$
Work Step by Step
Given $$ \int x^{3} \cos \left(x^{2}\right) d x$$Let
\begin{align*}
z =x^2 \ \ \ \ \ \ \ \ \ dz=2xdx
\end{align*}
Then
$$ \int x^{3} \cos \left(x^{2}\right) d x=\frac{1}{2}\int z \cos \left(z\right) dz$$
Let
\begin{align*}
u&=z \ \ \ \ \ dv = \cos \left(z\right) d z\\
du&=dz \ \ \ \ \ v = \sin \left(z\right)
\end{align*}
Then
\begin{align*}
\int x^{3} \cos \left(x^{2}\right) d x&=\frac{1}{2}\int z \cos \left(z\right) dz\\
&=\frac{1}{2} \left( z\sin z -\int \sin zdz\right)\\
&= \frac{1}{2} \left( z\sin z+ \cos z\right)+C\\
&= \frac{1}{2}\left(x^{2} \sin x^{2}+\cos x^{2}\right)+C
\end{align*}
