How to prove this? $$\int_0^{\pi/2}\cos^m(x)\cos (nx)\,dx=\frac m{m+n}\int_0^{\pi/2}\cos^{m-1}(x)\cos((n-1)x)\,dx.$$
I know that need to use this formula $\cos(A-B)$ and integration by parts. But I have no idea how to start as the reduction formula is for $\sin^m(x)$ and $\cos^n(x)$ and not $\cos^m(x)$ and $\cos (nx)$
Any help is appreciated.
Thank you in advance.
Use integration by parts. Write $\cos(n-1)x=\cos nx\cos x+\sin nx\sin x$ and define$$I:=\int_0^{\pi/2}\cos nx\cos^mxdx,\,u:=\sin nx,\,v:=-\frac1m\cos^mx$$so$$u^\prime=n\cos nx,\,v^\prime=\cos^{m-1}x\sin x,\,[uv]_0^{\pi/2}=0$$and$$\begin{align}\int_0^{\pi/2}\cos^{m-1}x\cos(n-1)xdx&=\int_0^{\pi/2}\cos^{m-1}x\cos nx\cos xdx\\&+\int_0^{\pi/2}\cos^{m-1}x\sin nx\sin xdx\\&=I+\int_0^{\pi/2}uv^\prime dx\\&=I-\int_0^{\pi/2}u^\prime vdx\\&=I+\frac{nI}{m}\\&=\frac{m+n}{m}I.\end{align}$$