$f(x)=\int_0^{\frac{\pi}{2}} e^{\sqrt{1-x^2 \sin^2 t}}\, dt$
$u=\sin t$
$f(x)=\int_0^{1} \cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$f'(x)=\int_0^{1} \frac{-xu^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$xf'(x)=\int_0^{1} \frac{-1+1-x^2u^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$xf'(x)=-\int_0^{1} \frac{1}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du+\int_0^{1} \sqrt{1-x^2 u^2 }\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$h(x)=\int_0^{1} \sqrt{1-x^2 u^2 }\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$h'(x)=\int_0^{1} \frac{-xu^2}{\sqrt{1-x^2 u^2 }}\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du-\int_0^{1} xu^2\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
$h'(x)=f'(x)-\int_0^{1} xu^2\cfrac{e^{\sqrt{1-x^2 u^2}}}{\sqrt{1-u^2}}\, du$
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