I'm trying to integrate this function with power series
$$\int2\cos^2(x)e^{-x}\,\mathrm dx$$
Since I know the Maclaurin series for $\cos(x)$ is $\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n}}{(2n)!}$ and $e^x$ is $\sum_{n=0}^{\infty} \dfrac{x^{n}}{n!}$ then
$$\int2\cos^2(x)e^{-x}\,\mathrm dx = \int2\left(\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n}}{(2n)!}\right)^2 \left(\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{n}}{n!}\right)\,\mathrm dx$$
However, I'm not certain that this is leading me forward. Usually these types of problems in early calculus involve representing the function under one sum and integrating the series as a polynomial. Is there something that I'm missing?
Hint: $$\cos^2x=\frac{1+\cos 2x}{2}$$ $$\cos(2x)e^{-x}= \left(\sum_{n=0}^{\infty} \dfrac{(-1)^n2^{2n}x^{2n}}{(2n)!}\right) \left(\sum_{m=0}^{\infty} \dfrac{(-1)^mx^{m}}{m!}\right)=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(-1)^{n+m}2^{2n}x^{2n+m}}{(2n)!m!}{}$$