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.

Hint:
$$(\sin(\cos(x)))'=-\sin(x)\cos(\cos(x)),$$
$$(\sin(\sin(x)))'=\cos(x)\cos(\sin(x)).$$
