Answer
$$-\frac{1}{2\ln 5} 5^{-2\sin x}+c
$$
Work Step by Step
Let $u=\sin x$, then $du=\cos x \ dx$. We have
$$\int \cos x \ 5^{-2\sin x}dx=\int 5^{-2u}du=-\frac{1}{2\ln 5}\int (-2\ln 5) \ 5^{-2u}du\\
=-\frac{1}{2\ln 5} 5^{-2u}+c=-\frac{1}{2\ln 5} 5^{-2\sin x}+c
$$
