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$$f(x)=\int_0^{\frac{\pi}{2}} e^{\sqrt{1-x^2 \sin^2 t}}\, dt$$

$u=\sin t$

$$f(x)=\int_0^{1} \cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$f'(x)=\int_0^{1} \frac{-xu^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$xf'(x)=\int_0^{1} \frac{-1+1-x^2u^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$xf'(x)=-\int_0^{1} \frac{1}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du+\int_0^{1} \sqrt{1-x^2 u^2 }\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$h(x)=\int_0^{1} \sqrt{1-x^2 u^2 }\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$h'(x)=\int_0^{1} \frac{-xu^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du-\int_0^{1} xu^2\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

$$h'(x)=f'(x)-\int_0^{1} xu^2\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$$

I am stuck after that. I have not found a relation.

Is it possible to find second order differential equation or higher order to satisfy $f(x)$? Can we express $f(x)$ in closed form as known functions? (Note: I especially try to express it as Complete elliptic integrals if it is possible.)

Thanks a lot for answers

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If $x$ not depends of $t$, substitution $u=x \sin{t}$ looks simpler – M. Strochyk Oct 4 '12 at 15:05
I'm guessing $|x| \le 1$ right? – Pragabhava Oct 4 '12 at 15:08
I don't think this integral is in a form which you can hope for a solution in terms of elliptic integrals. – Fabian Oct 4 '12 at 15:46
Special case $x=1$ is $\int_0^{\pi/2} e^{\cos t}dt$. But in closed form I only know this one: $\int_0^{\pi} e^{\cos t}dt = \pi I_0(1)$ with a Bessel function. – GEdgar Oct 4 '12 at 18:34

I have found no closed form for integrals of this type in the 10 years I did research in fast ways of evaluating diffraction integrals in imaging problems. If you need an analytical expression that avoids evaluating an integral in code, I have found Pade approximants useful so long as they respect branch points near the zero of the square root.

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