Answer
$$\int\sin x\sin(\cos x)dx=\cos(\cos x)+C$$
Work Step by Step
$$A=\int\sin x\sin(\cos x)dx$$
Let $u=\cos x$.
Then $du=-\sin xdx$. So $\sin xdx=-du$
Substitute into $A$, we have $$A=-\int\sin udu$$ $$A=-(-\cos u)+C$$ $$A=\cos u+C$$ $$A=\cos(\cos x)+C$$