$$\int_{1}^{1+2\pi}\cos(x)e^{(-\sin^2(x))}dx$$
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.
Best Answer
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 $0$.