Is there a closed form evaluation for

ò₀^p/2 xcos²ⁿ(x) 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 ò₀^p/2xcos²ⁿ(x) dx+ò₀^p/2 xsin²ⁿ(x) dx seems to have a connection to:

ò₀^p/2cos²ⁿ(x) dx which equals to:

ò₀^p/2sin²ⁿ(x) dx
although I can't pin-point it yet,

Yatir

It appears that (from your work, Yatir)

ò₀^p/2x(cos²ⁿx+sin²ⁿx)dx=((2n)!/n!²)4^-n-1p², 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

Also, for the original integral I_n=ò₀^p/2xcos²ⁿx dx you can derive a reduction formula.

Write cos²ⁿx=cos^2n-2x-cos^2n-2xsin² x. So:

I_n=I_n-1-ò₀^p/2 xcos^2n-2xsin² x dx

Integrate the second term by parts taking u=xsinx and dv/dx=cos^2n-2 x sinx, i.e. v=-cos^2n-1x/(2n-1). This gives

I_n=I_n-1-1/(2n×(2n-1))-I_n/(2n-1)

i.e.

2n I_n=(2n-1)I_n-1-1/(2n)