Computing of $\int_{-1}^1\frac{e^{ax}dx}{\sqrt{1-x^2}}, \: a \in \mathbb{R}$ I would like to find Fourier series for $f(x) = e^{ax}$ using Chebyshev polynomials. And first step is computation following integral.
How to compute $$\int_{-1}^1\frac{e^{ax}dx}{\sqrt{1-x^2}}, \: a \in \mathbb{R}$$

By setting $x=\sin\theta$ we have
$$ I(a)=\int_{-1}^{1}\frac{e^{ax}}{\sqrt{1-x^2}}\,dx = \int_{-\pi/2}^{\pi/2}\exp\left(a\sin\theta\right)\,d\theta \tag{1}$$
and we may expand the exponential function as its Taylor series at the origin. Since the integral of and odd integrable function (like $\sin^3$ or $\sin^5$) over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is zero, we get:
$$\begin{eqnarray*} I(a) &=& \sum_{n\geq 0}\frac{a^{2n}}{(2n)!}\int_{-\pi/2}^{\pi/2}\sin^{2n}(\theta)\,d\theta\\ (2i \sin\theta=e^{i\theta}-e^{-i\theta})\qquad &=&\sum_{n\geq 0}\frac{\pi a^{2n}}{4^n (2n)!}\binom{2n}{n}\\&=&\pi\sum_{n\geq 0}\frac{a^{2n}}{4^n(n!)^2}=\color{red}{\pi\cdot I_0(a)}.\end{eqnarray*} \tag{2}$$
