Integrate $\sin(\cos x)\text{d}x$ Soo.. a question appeared in my exam in which I had to compare the values of the following integrals $$J=\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx\\I=\int_0^{\pi/2} \cos(\sin x)\,\mathrm dx \\K=\int_0^{\pi/2} \cos x\,\mathrm dx$$ I was able solve it using an indirect method but it left me wondering whether I could find the exact values of the three integrals. 
PS: I have linked a picture of the problem and its solution for your reference. 

 A: $$J=\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx$$
This integral cannot be expressed in terms of a finite number of elementary functions. It can be written on the form of an infinite series or with a special function : the Struve function.
Change of variable : $t=\cos(x)$
$$J=\int_0^1 \frac{\sin(t)}{\sqrt{1-t^2}}\,\mathrm dx$$
See Eq.$(2)$ in http://mathworld.wolfram.com/StruveFunction.html
$\int_0^1 \sin(zt)(1-t^2)^{\nu-\frac12}\,\mathrm dx = \frac{\Gamma(\nu+\frac12)\Gamma(\frac12)}{2(z/2)^\nu}\mathrm H_\nu(z)$
With $\nu=0$ and $z=1$ :
$\int_0^1 \sin(t)(1-t^2)^{-\frac12}\,\mathrm dx = \frac{\Gamma(\frac12)^2}{2}\mathrm H_0(1) = \frac{\pi}{2}\mathrm H_0(1) $
$$\int_0^{\pi/2} \sin(\cos x)\,\mathrm dx=\frac{\pi}{2}\mathrm H_0(1)$$
$\mathrm H_0(1)=\sum_{k=0}^\infty \frac{(-1)^k}{2^{2k+1}\left(\Gamma(k+\frac32)\right)^2}$
A similar calculus leads to 
$$\int_0^{\pi/2} \cos(\sin x)\,\mathrm dx=\frac{\pi}{2}\mathrm J_0(1)$$
where $\mathrm J_0$ is the Bessel function of the first kind and order $0$.
$\mathrm J_0(1)=\sum_{k=0}^\infty \frac{(-1)^k}{2^{2k+1}\left(k!\right)^2}$
http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html
