Definite Integral (5)

Consider the general one:
I_n=\int_0^{\pi/2} \cos^nx\cos(n+1)x\ \mathrm dx
Using integration by parts, we get
I_n =\left. \cos^n x\dfrac{\sin(n+1)x}{n+1}\right|_0^{\pi/2}+\dfrac{n}{n+1}\int_0^{\pi/2} \cos^{n-1}x \sin x \sin(n+1)x\ \mathrm dx
The first term evaluates to zero. Writing
sin(n+1)x = sin nx cos x + cos nx sin x, we get
\cos^{n-1}x \sin x \sin (n+1)x = \cos^{n-1} x \cos nx - \cos^n x\cos (n+1)x
And so we get
I_n = \dfrac{n}{n+1}I_{n-1} -\dfrac{n}{n+1}I_n
giving
\boxed{I_n = \dfrac{n}{2n+1}\,I_{n-1}\quad \left(n\ne -\frac{1}{2}\right)}
From here, its easy enough (try it!) to obtain
\boxed{I_n =\dfrac{2^n(n!)^2}{(2n+1)!}}