Was trying to solve this: $C_r=\frac{2}{\pi}\int_{-1}^1\frac{e^xT_r(x)}{\sqrt{1-x^2}}dx$
where $r=0,1,2,3$
$T_r(x) =cosr[{cos}^{-1}x]$
While solving, I equated $x=cos\theta$
Therefore $T_r(x)=cosr\theta$
and $dx=-\sqrt{1-x^2}d\theta$ $$C_r=\frac{2}{\pi}\int_{\pi}^0\frac{e^{cos\theta}cosr\theta}{\sqrt{1-x^2}}.-\sqrt{1-x^2}d\theta$$
Solving further $$C_r=\frac{2}{\pi}\int_0^{\pi}e^{cos\theta}cosr\theta d\theta$$
When $r=0$, $C_r=\frac{2}{\pi}\int_0^{\pi}e^{cos\theta} d\theta$
When $r=1$, $C_r=\frac{2}{\pi}\int_0^{\pi}e^{cos\theta}cos\theta d\theta$
And so on. The problem is to now calculate the definite integrals without resorting to approximations using Trapezoid or Simpsons rule.