Question: > 
> Evaluate $\int \cos(\cos x)~dx$
> 
> 
> 

I tried to use chain rule but failed. Can anyone help me please?

Answer: 
This integral doesn't have a nice closed-form solution in terms of elementary functions, so this question is impossible (assuming you're just supposed to find the antiderivative in a form simpler than $\int \cos(\cos(x)) dx$)
$\int\cos\cos x~dx=\int\sum\limits\_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}~dx=\int\left(1+\sum\limits\_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$

For $n$ is any natural number,

$\int\cos^{2n}x~dx=\dfrac{(2n)!x}{4^n(n!)^2}+\sum\limits\_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

$\therefore\int\left(1+\sum\limits\_{n=1}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n)!}\right)~dx$

$=x+\sum\limits\_{n=1}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits\_{n=1}^\infty\sum\limits\_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$

$=\sum\limits\_{n=0}^\infty\dfrac{x}{4^n(n!)^2}+\sum\limits\_{n=1}^\infty\sum\limits\_{k=1}^n\dfrac{(-1)^n((k-1)!)^2\sin x\cos^{2k-1}x}{4^{n-k+1}(n!)^2(2k-1)!}+C$