An integral involving trigonometric functions and its inverse I have tried to evaluate the following integral for the last few hours, and I did not succeed:
$$ \int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arcsin}(r \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arcsin}(r \cdot \mathrm{cos}(\theta))} d \; \theta$$
for $0<r<1$. And also this other integral:
$$ \int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arctan}(t \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arctan}(t \cdot \mathrm{cos}(\theta))} d \; \theta.$$
Here $m$ and $n$ are integers,  and $t \in \mathbb{R}$ is scalar.
I am pretty sure that is nonzero, if and only if $n=m$, and indepedent of $r$ otherwise, but I cannot figure what substitution makes this easy to see.
 A: This addresses the question in its original form, where $\arcsin$ was used.
First, let's massage the integral:
$$ \begin{eqnarray}
   \mathcal{I} &=& \int_0^{2 \pi} \exp\left( i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta  \\
    &\stackrel{\theta \to 2\pi - \theta}{=}& 
    \int_0^{2 \pi} \exp\left( -i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta 
\end{eqnarray}
$$
Now averaging out both lines:
$$ \begin{eqnarray}
   \mathcal{I}
     &=& \int_0^{2 \pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \exp\left(  i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta  \\
    &=& 2 \int_0^{\pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \cos\left( m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta
\end{eqnarray}
$$
Here is a counter-example to your claim. Let $r = \frac{1}{2}$, and $n=1$, $m=2$. Then the integrand is positive, and hence the integral does not vanish:

Added The above counterexample actually carries over to the case with $\arctan$ just the same, i.e. the integrand is positive.
