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June 22nd, 2015, 09:05 PM   #1
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integral of exponential of sin squared

Compute the integral $\displaystyle\int_0^{\frac\pi 2} e^{\sin^2 x}dx$. When I tried, I made a careless (and wrong) assumption and got $\displaystyle\frac{\pi} {2}\sqrt e$. However, Wolfram Alpha gives my answer multiplied to the constant $I_0(1/2)$ where $I_n(z)$ is apparently a "Bessel function of the first kind."

I am curious if anyone can show me how to derive this result.
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June 24th, 2015, 04:55 AM   #2
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Substitute $\sin(x)^2$ with $\frac{1-\cos(2x)}{2}$.
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June 26th, 2015, 07:37 AM   #3
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Quote:
Originally Posted by ZardoZ View Post
Substitute $\sin(x)^2$ with $\frac{1-\cos(2x)}{2}$.
I have been at it for a while, but I don't see the connection to the Bessel differential equation.

EDIT: I may have got the Bessel differential equation wrong in the first place; I thought that $I_0(x)$ was the solution to $xy'' + y' +xy=0$ (although I don't know the initial conditions).
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