Integral $\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$ Can you please help me with this integral
$$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$
Thanks!

$$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$
Substitution:$x=\sin\theta$
$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta$$
$$e^{\sin\theta}=1+\frac{\sin\theta}{1!}+\frac{\sin^2\theta}{2!}+\frac{\sin^3\theta}{3!}.......$$
The odd powers of sine when used in the integral will produce zero.
Again for even values of n we have
  $$\int_{-\pi/2}^{\pi/2}\sin^n\theta d\theta=2 \times \frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}$$
Therefore,
$$I=\int_{-\pi/2}^{\pi/2}e^{\sin\theta}d\theta=\pi+2\times\Sigma\frac{1}{n!}[\frac{n-1}{n}\frac{n-3}{n-2}\frac{n-5}{n-4}....\frac{3}{4}\frac{1}{2}\frac{\pi}{2}]$$
$$=\pi+\pi\Sigma [\frac{1}{n(n-2)(n-4)......2}]^2$$
["n" running from two through even values ]
Writing n=2m we have,
$$\pi&ltI=\pi\Sigma[\frac{1}{2^m}\frac{1}{m!}]^2&lt4$$
[m running from  0 to infinity]
