I found the following on Wolfram Math World:
I understand the first step, the integration by parts. For the first case, m is even, I am pretty sure it involves a binomial expansion of $(\frac{1}{2}(1-\cos(2x)))^n$, which arises from writing $\cos^{2n}(x)$ as $(cos^2(x))^n$, then writing $cos^2(x)$ as $\frac{1}{2}(1-cos(2x)$, but i do not know how they integrated from there.
Here is the link for reference: http://mathworld.wolfram.com/CosineIntegral.html
For the case $m\equiv 2n$, one defines $$J_n=\int\cos(x)^{2n}dx$$ Which was already shown to be $$\begin{align} J_n&=\frac1{2n}\sin(x)\cos(x)^{2n-1}+\frac{2n-1}{2n}\int\cos(x)^{2n-2}dx\\ \therefore J_n&=\frac1{2n}\sin(x)\cos(x)^{2n-1}+\frac{2n-1}{2n}J_{n-1} \end{align}$$ And we may show from induction (see here) that if $$f_n=\alpha_n+\beta_nf_{n-1}$$ Then $$f_n=f_0\prod_{k=1}^{n}\beta_k+\sum_{k=0}^{n-1}\alpha_{n-k}\prod_{j=1}^{k}\beta_{n-j+1}$$ Setting $$f_n=J_n\qquad n\geq0$$ $$\alpha_n=\frac1{2n}\sin(x)\cos(x)^{2n-1}\qquad n\geq1$$ $$\beta_n=\frac{2n-1}{2n}\qquad n\geq1$$ $$f_0=J_0=\int dx=x$$ Gives the desired result.
