Double substitution when integrating. I need to integrate
$$f(x) = \cos(\sin x)$$
my first thought was substituting so that $u = \cos(x)$, but that doesn't seem to do the trick. Is there any way to do a double substitution on this?
Any ways on how to proceed would be appreciated.
 A: Although the indefinite integral $($or anti-derivative$)$ cannot be expressed in terms of elementary functions, its definite counterpart does have a closed form in terms of the special Bessel function:
$$\int_0^\tfrac\pi2\cos(\sin x)~dx~=~\int_0^\tfrac\pi2\cos(\cos x)~dx~=~\frac\pi2J_0(1).$$
Also, $$\int_0^\tfrac\pi2\sin(\sin x)~dx~=~\int_0^\tfrac\pi2\sin(\cos x)~dx~=~\frac\pi2H_0(1),$$
where H represents the special Struve function.
A: I tried to use WolframAlpha but no standard integral found.
So, you can approximate the function to an easier function.
cos(sin(x)) = a cos(bx)+c
if a = 0.25, b = 2, c = 0.75
then the approximation is almost inline with the function (I used a plotting program for this conclusion).
$$
\int f(x)dx = \int 0.25\cos(2x)+0.75\ dx \\= 0.5\sin(2x)\ +0.75x+C
$$
A: Or you also can solve it numerically by expand the function (e.g. use Taylor expansion) 
and then integrate each of its terms (e.g. first 5 terms).
A: You can express in terms of Incomplete Bessel Functions:
Consider $$\begin{align}J_0(1,w)&=\dfrac{2}{\pi}\int_0^w\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\cos\left(x-\dfrac{\pi}{2}\right)~d\left(x-\dfrac{\pi}{2}\right)
\\&=\dfrac{2}{\pi}\int_\frac{\pi}{2}^{w+\frac{\pi}{2}}\cos\sin x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\sin x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_\frac{\pi}{2}^0\cos\sin\left(\dfrac{\pi}{2}-x\right)~d\left(\dfrac{\pi}{2}-x\right)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{2}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\pi^\frac{\pi}{2}\cos\cos(\pi-x)~d(\pi-x)
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\frac{\pi}{2}\cos\cos x~dx-\dfrac{1}{\pi}\int_\frac{\pi}{2}^\pi\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-\dfrac{1}{\pi}\int_0^\pi\cos\cos x~dx
\\&=\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx-J_0(1)\end{align}
$$
Then $\dfrac{2}{\pi}\int_0^{w+\frac{\pi}{2}}\cos\sin x~dx=J_0(1)+J_0(1,w)$
$\therefore\int\cos\sin x~dx=\dfrac{\pi}{2}\left(J_0(1)+J_0\left(1,x-\dfrac{\pi}{2}\right)\right)+C$
