How to solve this definite integral? (Substitution method didn't help) 
$$\int_{1}^{1+2\pi}\cos(x)e^{(-\sin^2(x))}dx$$

an intuitive substitution would be $t = \sin(x)$ so we get $dt = \cos(x)dx$ and we're done with the $\cos(x)$, and then the range of integration would be from $\sin(1)$ to $\sin(1)$ which means we get $0$ as answer. But that doesn't satisfy the substitution method rules, because $t = \sin(x)\to x = \arcsin(t)$ and the output of this function is in the range $[\frac{-\pi}{2},\frac{\pi}{2}]\nsubseteq [1,1+2\pi]$, so we're out of range. 
I'm really wondering how to correctly solve this integral, any help or push in the right direction would be appreciated, thanks in advance.
 A: Expand the integral as
$$\int_1^{1+2\pi} \cos(x) e^{-\sin^2(x)} \, dx = \left\{\int_0^{2\pi} - \int_0^1 + \int_{2\pi}^{1+2\pi} \right\} \cos(x) e^{-\sin^2(x)} \, dx$$
For the first integral, we have
$$\begin{align*}
\int_0^{2\pi} \cos(x) e^{-\sin^2(x)} \, dx &= \left\{\int_0^\pi + \int_\pi^{2\pi}\right\} \cos(x) e^{-\sin^2(x)} \, dx \\[1ex]
&= \int_0^\pi \cos(x) e^{-\sin^2(x)} \, dx - \int_0^\pi \cos(x) e^{-\sin^2(x)} \, dx \\[1ex]
&=0
\end{align*}$$
where the integral over $[\pi,2\pi]$ is done by substituting $x\mapsto x+\pi$. In other words, let $x = y + \pi$, so $dx=dy$ and $x\in[\pi,2\pi]\implies y\in[0,\pi]$. Then
$$\begin{align*}
\int_\pi^{2\pi} \cos(x) e^{-\sin^2(x)} \, dx &= \int_0^\pi \cos(y + \pi) e^{-\sin^2(y+\pi)} \, dy & x \mapsto y+\pi \\[1ex]
&= \int_0^\pi -\cos(y) e^{-(-\sin(y))^2} \, dy & {\cos(y+\pi)=-\cos(y),\\\sin(y+\pi) = -\sin(y)}\\[1ex]
&= - \int_0^\pi \cos(x) e^{-\sin^2(x)} \, dx & y\mapsto x
\end{align*}$$
The integral over $[2\pi,1+2\pi]$ is equivalent to the integral over $[0,1]$ (easily shown by substituting $x\mapsto x-2\pi$), so the overall result is $0$.
A: One way to solve it is to note the function is odd about $x=1+{\pi}$. Hence the integral(which is essential signed area) , is odd
