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