From this answer:

$$\int_{0}^{2\pi} e^{\sin t} = \int_{0}^{\pi} e^{\sin t}\ dt+\int_{0}^{\pi} e^{-\sin t}\ dt$$

How to prove this?

My attempt:

$e^{-\sin t}$ is neither odd, nor even; the usual way to split such an integral could be

$$\int_{0}^{2\pi} e^{\sin t} = \int_{0}^{\pi} e^{\sin t}\ dt+\int_{\pi}^{2\pi} e^{\sin t}\ dt$$

The second addend could become, putting $u = -t$, $du = -dt$ and $\sin(-t) = -\sin(t)$:

$$- \int_{-2\pi}^{-\pi} e^{-\sin t}\ dt$$

but however it is not like in the initial expression.


The values of sinus function from $\pi$ to $2\pi$ are the same as minus the sinus function from $0$ to $\pi$. Or to be a bit more precise, consider a substitution $t\rightarrow t-\pi$.

Now what is sin$(t-\pi)$ ?


You have that $\sin(t)=-\sin (t-\pi)$. Apply to the integral over $(\pi,2\pi)$ and change variables $s=t-\pi$.


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