What is the integral of $\sin(\cos x)$? I've been wondering, what is the general form for the integral of $\sin(\cos x)$? I tried integrating it myself and couldn't, tried with Wolfram Alpha and it couldn't either. If you think about it conceptually, it should have a general form as it's graph if just like sin and cos, which both have general forms. If it doesn't have a general form, I'd like to know why. Thanks!
 A: It does not have an elementary antiderivative.  Calculus classes can give you an erroneous impression that you can always write down a formula for an antiderivative of a "nice" function.  This is not the case.
A: You can also try this one if you want.
Expanding $\sin{\cos {x}}$ in Taylor series expansion.
$\sin{\cos {x}}$ = $\cos {x}$ - $\cos^3 {x}$/3! + $\cos^5 {x}$/5! - $\cos^7 {x}$/7!+  -.....
Then one can integrate term by term.
There is no closed form solution exists for this function as Lucian suggested.
A: $\int\sin(\cos x)~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n+1}x}{(2n+1)!}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\cos^{2n}x}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n(1-\sin^2x)^n}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n(-1)^k\sin^{2k}x}{(2n+1)!}d(\sin x)$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k}x}{(2n+1)!k!(n-k)!}d(\sin x)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}n!\sin^{2k+1}x}{(2n+1)!k!(n-k)!(2k+1)}+C$
A: 
What is the integral of $\sin(\cos x)$ ?

So glad you asked ! :-) Although the indefinite integral does not possess a closed form, its definite counterpart can be expressed in terms of certain special functions, such as Struve H and Bessel J.
$$\int_0^\tfrac\pi2\sin(\sin x)~dx=\int_0^\tfrac\pi2\sin(\cos x)~dx=\frac\pi2H_0(1)$$
$$\int_0^\tfrac\pi2\cos(\sin x)~dx=\int_0^\tfrac\pi2\cos(\cos x)~dx=\frac\pi2J_0(1)$$
Also, $$\int_0^\tfrac\pi2\cos(\tan x)~dx=\int_0^\tfrac\pi2\cos(\cot x)~dx=\frac\pi{2e}$$
Hope this helps ! :-)
