I have some questions for double integrals, please help.... I really need help $$\int_{0}^{2\pi}\int_{1}^{e}\sin^{2009}(x)\frac{1}{y}dydx$$
In this question am I allowed to just pulled out the $2\pi$ right away from the integral from $0$ to $2\pi$? 
So that it becomes:
$$2\pi\int_{1}^{e}\sin^{2009}(x)\frac{1}{y}dydx$$

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{e}\frac{\sin(y)}{x}dxdy$$
Could I do the same in this question by pulling out the $\pi/2$ right away for this question? If not, why can't I do it for this question? Can't I do it whenever there is no theta anywhere and the integral contains $\pi$?
Thank you
 A: Change the order of integration. So first integrate $\sin^{2009}(x)\,dx$ from $0$ to $2\pi$. The geometry should tell you the answer is very simple.  Think of what the graph of $\sin x$ looks like from $0$ to $2\pi$. Or maybe ask a graphing program to graph $\sin^3 x$.
If you want to do the thing formally, break up the integral at $\pi$. Don't touch the first integral, $0$ to $\pi$. For the second integral, make the change of variable $x=t+\pi$. Then use the fact that $\sin(\pi +\theta)=-\sin\theta$.
The second integral is a little puzzling, since for the $\frac{1}{x}$ part, there is a problem at $0$. The improper integral does not exist.
A: You do not need to integrate. You just use the fact that if $y=f(x)$ is an odd function in $[-a,a]$, then
$$ \int_{-a}^af(x)dx=0. $$
Note that $y=\sin^{2009}x$ is an odd function in $[-\pi,\pi]$. Now
\begin{eqnarray*}
\int_0^{2π}\int^e_1\sin^{2009}(x)\frac{1}{y}dydx&=&\int_0^{2π}\sin^{2009}(x)dx\\
&=&-\int_{-\pi}^{\pi}\sin^{2009}(x)dx\\
&=&0.
\end{eqnarray*}
