# To find closed form of $f(x)=\int_0^{\frac{\pi}{2}} e^{\sqrt{1-x^2 \sin^2 t}}\, dt$ as known functions

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

$u=\sin t$

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

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

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

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

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

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

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

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.)

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
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