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!
 A: $$\int_{-1}^1 \frac{e^x}{\sqrt{1-x^2}}dx$$
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]
A: Let me try to sort out the previous takes. For this, we will have to take as a prerequisite the series for the zeroth-order modified Bessel function of the first kind
$$I_0(x)=\sum_{k=0}^\infty \frac1{(k!)^2}\left(\frac{x^2}{4}\right)^k$$
which can be derived from the modified Bessel differential equation
$$x^2 y^{\prime\prime}+xy^\prime=x^2 y$$
with initial conditions $y(0)=1,y^\prime(0)=0$.
Now,
$$\begin{align*}
\int_{-1}^1 \frac{\exp\,x}{\sqrt{1-x^2}}\mathrm dx&=\sum_{k=0}^\infty \frac1{k!}\int_{-1}^1 \frac{x^k}{\sqrt{1-x^2}}\mathrm dx\\
&=\sum_{k=0}^\infty \frac1{(2k)!}\int_{-1}^1 \frac{(x^2)^k}{\sqrt{1-x^2}}\mathrm dx\\
&=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^1 \frac{(x^2)^k}{\sqrt{1-x^2}}\mathrm dx\\
&=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^{\pi/2} \frac{\sin^{2k} u \; \cos\,u}{\sqrt{1-\sin^2 u}}\mathrm du\\
&=2\sum_{k=0}^\infty \frac1{(2k)!}\int_0^{\pi/2} \sin^{2k} u\mathrm du\\
&=\pi\sum_{k=0}^\infty \frac1{(2k)!}\frac{(2k)!}{4^k (k!)^2}\\
&=\pi\sum_{k=0}^\infty \frac1{(k!)^2}\left(\frac14\right)^k=\pi I_0(1)\\
\end{align*}$$
where we used the odd/even properties of the moments in the second and third lines, and the Wallis formula in the fifth line.
