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So i'm trying to find out how, or if its even possible, to integrate $e^{\sin(x)\cos(x)}$ analytically from $0$ to $2\pi$.

I know that i can integrate $e^{\cos(x)+\sin(x)}$ or $e^{\cos(x)^2}$ and these just give me Bessel functions.. but i don't know how to do the one in the title.. and google searches are coming up short.

Does anyone have any idea?

Cheers

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The integral is $2\pi I_0(\frac12)$ where $I_0(x)$ is the modified bessel function of the first kind. Please look up its integral representation on the wiki page. –  achille hui Dec 27 '13 at 17:17
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2 Answers 2

You can use $\cos(x)\sin(x)=\frac 12 \sin(2x)$ and maybe $u=2x$ to get to a form you know.

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$\int_0^{2\pi}e^{\sin x\cos x}~dx$

$=\int_0^{2\pi}e^{\frac{\sin2x}{2}}~dx$

$=\dfrac{1}{2}\int_0^{4\pi}e^{\frac{\sin x}{2}}~dx$

$=\int_0^{2\pi}e^{\frac{\sin x}{2}}~dx$

$=\int_{-\frac{\pi}{2}}^\frac{3\pi}{2}e^{\frac{\sin\bigl(\frac{\pi}{2}+x\bigr)}{2}}~d\biggl(\dfrac{\pi}{2}+x\biggr)$

$=\int_{-\frac{\pi}{2}}^\frac{3\pi}{2}e^{\frac{\cos x}{2}}~dx$

$=\int_{-\pi}^\pi e^{\frac{\cos x}{2}}~dx$

$=\int_{-\pi}^0e^{\frac{\cos x}{2}}~dx+\int_0^\pi e^{\frac{\cos x}{2}}~dx$

$=\int_\pi^0e^{\frac{\cos(-x)}{2}}~d(-x)+\int_0^\pi e^{\frac{\cos x}{2}}~dx$

$=\int_0^\pi e^{\frac{\cos x}{2}}~dx+\int_0^\pi e^{\frac{\cos x}{2}}~dx$

$=2\int_0^\pi e^{\frac{\cos x}{2}}~dx$

$=2\pi I_0\left(\dfrac{1}{2}\right)$

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