Integral $\int_0^{2\pi}e^{\sin^2x}dx$ My attempt at a solution using the substitution method:
$$u=\sin^2x\\du=2\sin x\cos xdx\\du=2u^{\frac12}{(1-\sin^2x)}^{\frac12}dx\\du=2u^{\frac12}{(1-u)}^{\frac12}dx\\\int_0^{2\pi}e^{\sin^2x}dx\\\int_0^1e^{u}\frac{1}{2u^{\frac12}{(1-u)}^{\frac12}}du$$
I have no idea how to solve it.
thanks in advance.
 A: According to Wolfram, it is
$$\int_{0}^{2\pi} e^{[\sin x]^2}\ dx = 2\sqrt{e} \pi I_0\left(\frac{1}{2}\right)$$
I think there isn't really a good "trick" for this beyond just recognizing that the modified Bessel functions have well-known integral identities involving integrands of a broadly similar shape in that they involve $e^{\cos x}$, so one should then try to mash them to see if one can get something to come out. Namely,
$$I_\alpha(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} \cos \alpha \theta\ d\theta - \frac{\sin \alpha \pi}{\pi} \int_{0}^{\infty} e^{-x \cosh t - \alpha t}\ dt$$
is a "known quantity" in that it's published already, and note that if one takes $\alpha = 0$, the right-hand term dies and the left becomes
$$I_0(x) = \frac{1}{\pi} \int_{0}^{\pi} e^{x \cos \theta} d\theta$$
hence use the half-angle formula $[\sin x]^2 = \frac{1 - \cos 2x}{2}$ so
$$\int_{0}^{2\pi} e^{[\sin x]^2}\ dx = \int_{0}^{2\pi} e^{\frac{1 - \cos 2x}{2}}\ dx = \int_{0}^{2 \pi} e^{1/2} e^{-\frac{1}{2} \cos{2x}}\ dx = \sqrt{e} \int_{0}^{2\pi} e^{-\frac{1}{2} \cos 2x}\ dx$$
then you should be able to take it from there with scalings.
