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Welcome to NRICH.

ò₀ ^p/2 x cos²ⁿ(x) dx

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By Brad Rodgers on Wednesday, August 28, 2002 - 09:24 pm:

Is there a closed form evaluation for

ò₀ ^p/2 x cos²ⁿ(x) dx,

for integer n in general? I can easily evaluate it for individual n, but I don't see the pattern...

Brad

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By Yatir Halevi on Wednesday, August 28, 2002 - 11:01 pm:

I don't know if it helps,
but ò₀ ^p/2xcos²ⁿ(x)dx+ò₀ ^p/2xsin²ⁿ(x)dx seems to have a connection to:
ò₀ ^p/2cos²ⁿ(x)dx which equals to:
ò₀ ^p/2sin²ⁿ(x)dx

altough I can't pin-point it yet,

Yatir

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By Brad Rodgers on Thursday, August 29, 2002 - 01:58 am:

It appears that (from your work, Yatir)

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

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

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By Ian Short on Thursday, August 29, 2002 - 10:39 am:

Sorry to butt in- Possibly by substituting y=p/2-x into ò₀ ^p/2xsin²ⁿxdx it becomes clear how to evaluate the integral in the previous message.

Ian

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By Michael Doré on Thursday, August 29, 2002 - 04:37 pm:

Also, for the original integral I_n = ò₀ ^p/2 x cos²ⁿx dx you can derive a reduction formula.

Write cos²ⁿx = cos^2n-2x - cos^2n-2x sin²x. So:

I_n = I_n-1 - ò₀ ^p/2x cos^2n-2x sin²x dx

Integrate the second term by parts taking u = x sin x and dv/dx = cos^2n-2x sin x, 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)

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