I would like to integrate: $$\int_0^{1} \operatorname{e}^{\sqrt{1-x^2}}dx$$ I tried to make a change of variable $\sqrt{1-x^2}\rightarrow X$ but I get: $$\int_0^{1} \frac{e^X X}{\sqrt{1-X^2}} \, dX=[-e^X\sqrt{1-X^2}]_0^1+\int_0^{1} e^X\sqrt{1-X^2}dX $$ which is not easier...
Could you help me to do so please ?
If you let $x=\sin(t)$, you are left with $$\int_0^{1} e^{\sqrt{1-x^2}}\,dx=\int_0^{\frac \pi 2} e^{\cos(t)}\, \cos(t)\,dt$$ Now, have a look at my answer to this question and I am sure that you will percieve the complexity of your problem (notice that using the expansion, the two problems are identical). Taking into account the bounds, the exact result is $$\int_0^{1} e^{\sqrt{1-x^2}}\,dx=\frac{\sqrt{\pi }}2\sum_{n=0}^\infty \frac{ \Gamma \left(\frac{n}{2}+1\right)}{\Gamma(n+1) \,\Gamma \left(\frac{n+3}{2}\right)}=\frac \pi 2 (\pmb{L}_{-1}(1)+I_1(1))$$ where appear Struve and Bessel functions.
Now, you can make quite good approximations using Padé approximant for the integrand. A simple one (built around $x=0$) would be $$ e^{\sqrt{1-x^2}}\sim e \frac{13 x^4-54 x^2+48}{-2 x^4-30 x^2+48}$$ Write it a $$ e \frac{13 x^4-54 x^2+48}{-2 x^4-30 x^2+48}=-\frac {13e}2\frac {(x^2-a)(x^2-b)}{(x^2-c)(x^2-d)}$$ with $$a=\frac{27-\sqrt{105}}{13}\qquad b=\frac{27+\sqrt{105}}{13}\qquad c=-\frac{\sqrt{321}+15}{2}\qquad d=\frac{\sqrt{321}-15}{2}$$ and use partial fraction decomposition $$\frac {(x^2-a)(x^2-b)}{(x^2-c)(x^2-d)}=\frac{a b-a c-b c+c^2}{(c-d) (x^2-c)}+\frac{-a b+a d+b d-d^2}{(c-d) (x^2-d)}+1$$ and you will end with one arctangent and one hyperbolic arctangent.
The numerical result $$\int_0^1 e \frac{13 x^4-54 x^2+48}{-2 x^4-30 x^2+48}\,dx=2.24770$$ while numerical integration gives $$\int_0^1 e^{\sqrt{1-x^2}}\,dx=2.24395$$ that is to say a relative error of $0.17$%.