The definite integrals of $e^{-x} \sin^n{(x)}$ and $e^{-x} \cos^n{(x)}$ between $0$ and $\infty$ For the integral
$$\begin{align}S_n =\int_{0}^{\infty} e^{-x} \sin^n{(x)}dx\end{align}$$
I have found a formula for $S_n$:
$$\begin{align} S_n=2^{\frac{(-1)^n-1}{2}} n!! (n-1)!! \prod_{2\le k\le n}^{{k\equiv n \mod{2}}} \frac{1}{k^2+1} \end{align}$$
By solving the following recurrence relation:
$$\begin{align} S_n=\frac{n(n-1)}{n^2+1}S_{n-2} \end{align}$$
But I cannot find a similar general formula for the integral
$$C_n=\int_{0}^{\infty} e^{-x} \cos^n{(x)}dx$$
Given that I have found the following recurrence relation for $C_n$:
$$\begin{align} C_n = \frac{1}{n^2+1}+ \frac{n(n-1)}{n^2+1}C_{n-2}\end{align}$$
Is it possible to find a general formula for $C_n$ in a similar format to that of $S_n$?
 A: I'm not sure this will help, but using the fact that $\cos^n(x) = \displaystyle \frac{(e^{ix}+e^{-ix})^n}{2^n}$, we have
$$\begin{align*}
\int_0^\infty e^{-x}\cos^n(x)\mathrm d x &= \frac1 {2^n}\int_0^\infty e^{-x}\sum_{k=0}^n \binom n k e^{i(2k-n)x}\mathrm d x\\
&=\frac1 {2^n}\sum_{k=0}^n \binom n k\frac{1}{1+i(n-2k)}\\&=
\frac1 {2^{n+1}}\sum_{k=0}^n \binom n k\left(\frac{1}{1+i(n-2k)}+\frac{1}{1-i(n-2k)}\right)\\
&=\frac 1 {2^{n}}\sum_{k=0}^n \binom n k\frac{1}{1^2+(n-2k)^2}.
\end{align*}$$
A: Please, fasten your seatbelt before reading !
Using a CAS and defining
$$A_n=\frac{2^{n+2}}{\sqrt{\pi }\, n!}C_n$$ we have
$$A_n=\frac{\sqrt{\pi } \left(2 e^{\pi } (-1)^n+1+e^{2 \pi }\right) (\coth (\pi
   )-1)}{\Gamma \left(\frac{n+2-i}{2} \right) \Gamma \left(\frac{n+2+i}{2}
   \right)}-\, _3\tilde{F}_2\left(1,\frac{n+2-i}{2} ,\frac{n+2+i}{2}
   ;\frac{n+3}{2},\frac{n+4}{2};1\right) $$
where appears the regularized generalized hypergeometric function.
By magic, this produces for the $C_n$'s the sequence
$$\left\{\frac{1}{2},\frac{3}{5},\frac{2}{5},\frac{41}{85},\frac{9}{26},\frac{263}{62
   9},\frac{101}{325},\frac{15357}{40885},\frac{7597}{26650},\frac{284603}{825877},
   \frac{43116}{162565},\frac{38393473}{119752165}\right\}$$
As I wrote in a comment after @Song's answer, I have the vague feeling that this could write in terms of incomplete beta functions but I have not been able to do it.
Edit
Concerning the $S_n$, the formula is effectively simpler
$$S_n=\frac{e^{\pi /2} \pi   \left((-1)^n+e^{\pi }\right) (\coth (\pi )-1) \,n!}{2^{n+1}\Gamma \left(\frac{n+2-i}{2} \right) \Gamma \left(\frac{n+2+i}{2}
   \right)}$$
Concerning the $C_n$, starting from @Song's answer, I finally obtained the surprising
$$C_n=\frac{i^{n+1-i} } {2^{n+2}}\left(B_{-1}\left(-\frac{n-i}{2},n+1\right)-(-1)^i
   B_{-1}\left(-\frac{n+i}{2},n+1\right)\right)$$
