How would you use Fubini's theorem to solve $\int_{0}^{1}\int_{\arcsin(y)}^{\frac{\pi}{2}} e^{\cos(x)} \, dx \, dy$ I think I need to re-define the upper and lower limits, but in any case I end up with:
$\int e^{\cos(x)} \, dx$ which doesn't allow me to solve the equation.
 A: The region of integration is that bounded by the $x$-axis, the line $x=\frac{\pi}{2}$, and the curve $x=\arcsin y$, or equivalently on this region, $y=\sin x$. Thus, applying Fubini's theorem gives that your integral is equal to
$$\int_0^{\frac{\pi}{2}} \int_0^{\sin x} e^{\cos x}\, dy\, dx$$
You can readily integrate with respect to $y$, since $e^{\cos x}$ is constant with respect to $y$, and then the integration with respect to $x$ is a simple application of integration by substitution.
A: $$
\int_0^1 \int_{\arcsin(y)}^{\pi/2} e^{\cos(x)} \, dx \, dy
$$
"Solve the equation" is not the right usage here; "evaluate the expression" would be appropriate.  It's not an equation.
You have
$$
0 < \arcsin y < x < \frac \pi 2.
$$
Since the sine function strictly increases on the interval $[0,\pi/2]$, this implies
$$
0 < y < \sin x < 1.
$$
Thus for each value of $x\in[0,\pi/2]$, the other variable, $y$, goes from $0$ to $\sin x$.  So we have
$$
\int_0^{\pi/2} \int_0^{\sin x} e^{\cos x}\,dy\,dx = \int_0^{\pi/2} \left[ \vphantom{\frac11} ye^{\cos x} \right]_{y:=0}^{y:=\sin x} \, dx = \int_0^{\pi/2}  (\sin x) e^{\cos x} \,dx = \int_1^0 e^u\,du.
$$
