Sin²ⁿ x

By Brad Rodgers on Tuesday, November 06, 2001 - 02:04 am:

Is there a general form for

ò₀^2p(sin(x))²ⁿ dx
For integer n? I can come up with results for individual n using contour integration, but nothing useful for all n.

Brad

By Kerwin Hui on Tuesday, November 06, 2001 - 02:17 am:

Brad,

You can actually find a formula for all positive integer n. One way of doing it - recall sin x = (exp(i x)-exp(-i x))/2i. Now take the 2nth power and notice that ò₀^2p e^{i m x} dx=0 for all non-zero integral m. Another way - try integrate by parts twice to give a reduction formula.

Kerwin

By Brad Rodgers on Thursday, November 08, 2001 - 06:39 pm:

I see the way using complex numbers, I but I can't seem to get a reduction formula. Could you post your method?

Thanks,

Brad

By Michael Doré on Thursday, November 08, 2001 - 06:55 pm:

Write:

sin²ⁿ x=sin^2n-1 x sin x

Set u=sin^2n-1 x, dv/dx=sin x, i.e. v=-cos x. After applying integration by parts, i.e.

òu dv/dx dx=u v-òv du/dx dx

then use the result cos² x=1-sin² x.