Closed form solution for $\int_{0}^{2\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x dx$

Question

I am trying to calculate Fourier series coefficients (by hand) and the integrals I need to solve are of the following type

$$I(N,M,n,m)=\int_{0}^{2\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x dx,$$

in which $N,M,n,m \in \{0,1,2,3\}$. I tried to use WolframAlpha / Maple to come up with a general formula because doing all the $4^4$ cases would make it impossible for me to work with that list but both didn't give a result.

It would be great if there was a way to obtain a simple closed form solution. If that is not possible is there a way to get very accurate approximation for $I(N,M,n,m)$?

Answer 1

$$I(N,M,n,m)=\int_{0}^{\pi}|\cos^N x|\sin^M x\cos^n x\sin^m x dx \\+\int_{\pi}^{2\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x,dx$$ Set $u =x-\pi$

$$\int_{\pi}^{2\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x,dx\\=\int_{0}^{\pi}|\cos^N x||\sin^Mx |\cos^n (x+\pi)\sin^m (x+\pi),dx $$ So that $$I(N,M,n,m)=\int_{0}^{\pi}|\cos^N x|\sin^M x\cos^n x\sin^m x dx \\+(-1)^{n+m}\int_{0}^{\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x,dx$$

But $\cos(x+\pi) = -\cos x$ and $\sin(x+\pi) = -\sin x$ Hence,

\begin{split} I(N,M,n,m)&=&\int_{0}^{\pi}|\cos^N x|\sin^M x\cos^n x\sin^m x dx \\ &+&(-1)^{n+m}\int_{0}^{\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x,dx\\ &=&[1+(-1)^{n+m}]\int_{0}^{\pi}|\cos^N x||\sin^M x|\cos^n x\sin^m x,dx\\ &=&[1+(-1)^{n+m}]\int_{0}^{\frac{\pi}{2}}\cos^N x\sin^M x\cos^n x\sin^m x dx \\ &+&[1+(-1)^{n+m}]\int_{\frac{\pi}{2}}^{\pi} |\cos^N x||\sin^M x|\cos^n x\sin^m x dx \end{split}

Setting $ t= \pi-x$ and noticing that $\cos(\pi-x) = -\cos x$ and $\sin(\pi-x) = \sin x$ We get ,

$$\int_{\frac{\pi}{2}}^{\pi} |\cos^N x||\sin^M x|\cos^n x\sin^m x dx = (-1)^n\int^{\frac{\pi}{2}}_{0} \cos^N x\sin^M x \cos^n x\sin^m x dx$$ Therefore,

\begin{split} I(N,M,n,m)&=& [1+(-1)^{n+m}]\int_{0}^{\frac{\pi}{2}}\cos^N x\sin^M x\cos^n x\sin^m x dx \\ &+&[1+(-1)^{n+m}]\int_{\frac{\pi}{2}}^{\pi} |\cos^N x||\sin^M x|\cos^n x\sin^m x dx\\ &=& [1+(-1)^{n+m}][1+(-1)^{n}]\int_{0}^{\frac{\pi}{2}}\cos^N x\sin^M x\cos^n x\sin^m x dx \end{split} and $$ [1+(-1)^{n+m}][1+(-1)^{n}]= [1+(-1)^{m}+(-1)^{n}+(-1)^{n+m}]$$

Conclusion $$\bbox[yellow]{I(N,M,n,m) =[1+(-1)^{m}+(-1)^{n}+(-1)^{n+m}]\int_{0}^{\frac{\pi}{2}}\cos^{N+n} x\sin^{M+m} xdx }$$

Also note the following relation with Beta function and Gamma function \url{https://fr.wikipedia.org/wiki/Fonction_b%C3%AAta}.

$$\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} =B(x,y) = 2\int_{0}^{\frac{\pi}{2}}\cos^{2x-1} t\sin^{2y-1} tdt ,~~x>0,y>0$$

So that ,

$$\bbox[yellow]{I(N,M,n,m) =\frac{1}{2}[1+(-1)^{m}+(-1)^{n}+(-1)^{n+m}]B\left(\frac{M+m+1}{2},\frac{n+N+1}{2}\right)\\=\frac{1}{2}[1+(-1)^{m}+(-1)^{n}+(-1)^{n+m}] \frac{\Gamma\left(\frac{M+m+1}{2}\right)\Gamma\left(\frac{N+n+1}{2}\right)} {\Gamma\left(\frac{M+N+m+n+2}{2}\right)}}$$

Answer 2

Hint: An easy way is to use the following identity which is quite easy to prove

which is an easy consequence of Integration, trigonometry, gamma/beta functions