Htykuut

If $$I_{m,n} = int_{0}^{pi/2}cos^m x cos nx ~dx$$
prove that
$$I_{m,n} = frac{m(m-1)}{m^2-n^2}I_{m-2,n}$$

I have tried my level best to solve but failed. Please help me out with a detailed solution if possible.This is what I have done...

Let $f_{m,a}(x)=f_m=e^{ax}cos^mx$. Take the derivative of $f_{m}$,
begin{align*}
f'_{m}&=af_{m}-mf_{m-1}sin x.tag{1}
end{align*}
So, if we take $g_{m}=af_{m}color{magenta}+mf_{m-1}sin x$, the derivative of $g_{m}$ will cancel $color{blue}{sin x}$ out, i.e.
begin{align*}
g'_{m}&=af'_{m}hspace{7.3em}+mbig(f'_{m-1}sin xhspace{9.9em}+underbrace{f_{m-1}cos x}_{=f_{m}}big)\
g'_{m}&=a^2f_{m}-color{blue}{amf_{m-1}sin x}+color{blue}mbig(color{blue}{af_{m-1}sin x}-(m-1) f_{m-2}color{orange}{sin^2 x}+f_mbig)\
g'_{m}&=(a^2+m)f_m-m(m-1)f_{m-2}color{orange}{(1-cos^2 x)}\
g'_{m}&=(a^2+m^2)f_m-m(m-1)f_{m-2}.tag{2}
end{align*}
Now, we assume
$$J_{m,a}=int_0^{pi/2}e^{ax}cos^mx~mathrm dx=int_0^{pi/2}f_m~mathrm dx,$$
integrate $(2)$,
$$underbrace{big[g_mbig]_0^{pi/2}}_{=-a}=(a^2+m^2)J_{m,a}-m(m-1)J_{m-2,a}.tag{3}$$
Put $a=in$, notice $Rebig{J_{m,in}big}=I_{m,n}$, by considering the real part of $(3)$, we get the desired result

$$0=(m^2-n^2)I_{m,n}-m(m-1)I_{m-2,n}.$$