Answer
$$\frac{1}{2}\left(\sin t^2 -\frac{1}{3}\sin^3 t^2\right)+c$$
Work Step by Step
Given
$$\int t\cos^3t^2dt $$
Let $$u=t^2\ \ \ \Rightarrow \ \ du=2tdt$$
Then
\begin{align*}
\int t\cos^3t^2dt&=\frac{1}{2}\int \cos^3udu\\
&=\frac{1}{2}\int \cos^2u\cos u du\\
&=\frac{1}{2}\int (1-\sin^2u)\cos u du\\
&=\frac{1}{2}\left(\sin u -\frac{1}{3}\sin^3 u\right)+c\\
&=\frac{1}{2}\left(\sin t^2 -\frac{1}{3}\sin^3 t^2\right)+c
\end{align*}