So I have been doing some integrals of the form $\int \sin^{2m}x\cos^{2n}x\mathrm dx $ using de Moivre's theorem and writing the integrand in the form $\sum_{r}a_{r}\sin(rx)+b_{r}\cos(rx)$. And I am trying to evaluate the integral in the general case. Here is what I've done.
$$\begin{aligned}\sin^{2m}x\cos^{2n}x &=\left(\frac{z-1/z}{2i}\right)^{2m}\left(\frac{z+1/z}{2}\right)^{2n}\\ &= \frac{-1}{2^{2m+2n}}\left(\sum_{k=0}^{2m}(-1)^{k}{2m \choose k}z^{2m-2k}\right)\left(\sum_{j=0}^{2n}{2n\choose j}z^{2n-2r}\right)\end{aligned}$$
The coefficient of $z^{r}$ will be at $k+j=m+n-r/2$ and similarly can be modified for $z^{-r}$. I don't know how to proceed. Any hints are appreciated. Thanks.
What you wrote is correct but it is clear that you are entering the domain of the gaussian hypergeometric function.
Let $\cos^2(x)=t$ to make $$I_{m,n}=\int \sin ^{2 m}(x) \cos ^{2 n}(x)\,dx=-\frac{1}{2} \int (1-t)^{m-\frac{1}{2}} t^{n-\frac{1}{2}}\,dt$$ $$I_{m,n}=-\frac{t^{n+\frac{1}{2}} }{2 n+1}\,\, _2F_1\left(\frac{1-2m}{2},\frac{2n+1}{2};\frac{2n+3}{2};t\right)$$ So, your difficulties are not surprising.