Is there a closed form evaluation for

${\int }_0^{\pi /2}x{\cos }^{2n}(x)\mathrm{dx}$,
for integer n in general? I can easily evaluate it for individual n, but I don't see the pattern...

Brad

I don't know if it helps,
but ${\int }_0^{\pi /2}x{\cos }^{2n}(x)\mathrm{dx}+{\int }_0^{\pi /2}x{\sin }^{2n}(x)\mathrm{dx}$ seems to have a connection to:

${\int }_0^{\pi /2}{\cos }^{2n}(x)\mathrm{dx}$ which equals to:

${\int }_0^{\pi /2}{\sin }^{2n}(x)\mathrm{dx}$
although I can't pin-point it yet,

Yatir

It appears that (from your work, Yatir)

${\int }_0^{\pi /2}x({\cos }^{2n}x+{\sin }^{2n}x)\mathrm{dx}=((2n)!/n{!}^2)4^{-n-1}{\pi }^2$, and I'll have a go at proving this. There's obviously a link between this and the original integral, but in the latter, a rational term appears that I just can't seem to determine.

(Of course, if anyone proves this, and makes a decent conjecture about the rational term, it'd be fairly easy to prove by induction; the problem is making the decent conjecture)

Brad

Ian

$2n{I}_n=(2n-1)I_{n-1}-1/(2n)$