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 $S_n$?

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
$$\begin{align*}
\int_0^\infty e^{-x}\cos^n(x)\mathrm d x &= \frac1 {2^n}\int_0^\infty e^{-x}\sum_{k=0}^n \binom n k e^{i(2k-n)x}\mathrm d x\\
&=\frac1 {2^n}\sum_{k=0}^n \binom n k\frac{1}{1+i(n-2k)}\\&=
\frac1 {2^{n+1}}\sum_{k=0}^n \binom n k\left(\frac{1}{1+i(n-2k)}+\frac{1}{1-i(n-2k)}\right)\\
&=\frac 1 {2^{n}}\sum_{k=0}^n \binom n k\frac{1}{1^2+(n-2k)^2}.
\end{align*}$$
