Calculate a double integral Why does the following identity hold?
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(\theta-\phi))e^{i(\theta-\phi)} \, d\theta \, d\phi=2\pi\int_0^{2\pi}\arcsin(k\cdot \cos x)e^{ix} \, dx$$
where $k\in[-1,1]$ is a constant.
I'm trying to use substitution $x=\theta-\phi$ but it still needs some tricks. Thanks!
 A: \begin{align}
& \int_0^{2\pi} \left( \int_0^{2\pi}\arcsin(k\cdot \cos(\theta-\varphi))e^{i(\theta-\varphi)} \, d\theta\right) \, d\varphi \\[10pt]
= {} & \int_0^{2\pi} \int_{-\varphi}^{2\pi-\varphi} \quad \arcsin(k\cos x) e^{ix} \, dx \quad d\varphi \tag A \\[10pt]
= {} & \int_0^{2\pi} \underbrace{\int_0^{2\pi} \arcsin(k\cos x) e^{ix} \, dx }_{\Large\text{The variable $\varphi$ does not appear here.}} \,d\varphi \tag B \\[12pt]
& \text{Line $(A)$ is equal to line $(B)$ because of periodicity} \\
& \text{of the function being integrated.} \\[10pt]
= {} & \int_0^{2\pi} (\text{constant}) \, d\varphi
\end{align}
In this context $\text{“}$constant$\text{''}$ means not depending on $\varphi;$ it means anything that does not change as $\varphi$ goes from $0$ to $2\pi.$ Thus the absence of the variable $\varphi$ from the expression inside the outer integral is the essential fact here. The integral of a constant over an interval is the constant times the length of the interval. The length of this interval is $2\pi-0.$ So the integral is equal to
$$
2\pi\times \text{constant.}
$$
And recall that the $\text{“}$constant$\text{''}$ is $\displaystyle \int_0^{2\pi} \arcsin(k\cos x) e^{ix}\, dx.$
A: Substitution $x = \theta - \phi$, then 
$$ \frac{dx}{d\theta} = 1$$
$\theta = 0$ gives $x = -\phi$, $\theta = 2\pi$ gives $x = 2\pi - \phi$
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(\theta-\phi))e^{i(\theta-\phi)} \, d\theta \, d\phi=$$
$$\int_0^{2\pi}\int_{-\phi}^{2\pi - \phi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx \, d\phi=$$
$$\int_0^{2\pi}\int_0^{2\pi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx \, d\phi=$$
$$2\pi \int_0^{2\pi}\arcsin(k\cdot \cos(x))e^{i(x)} \, dx=$$
Where we used that the integrand is a function that is $2\pi$ periodic, to convert the boundary terms in the third integral. In the last step we used that the inner integral does not depend on $\phi$.
