integral of $\cos{x}\cos{(\sin{x})}dx$ I set $u=\cos{x}$, $du=-\sin{x}dx$, $dx=\frac{1}{-\sin{x}}$
$u\cos{(\sin{x})}dx$
$u [-\sin{(\sin{x})}\cos{x}(1)]+C$
$\frac{-(\sin{(\sin{x}})).(\cos{x})}{-sin{x}} + C$
tentative answer: $\sin{(\cos{x})} + C$
Another source says that the answer is $sin{(\sin{x})} + C$, and I wanted to know if and where I went wrong.

$$\int{\cos x\cos (\sin x)dx}=\int{\cos (\sin x)d\left( \sin x \right)}=\sin (\sin x)+C$$
