Sin²ⁿ x

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

Is there a general form for

\int_{0}^{2 π} (\sin (x {))}^{2 n} 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 $\sin x = (\exp (i x) - \exp (- i x)) / 2 i$ . Now take the $2 n$ th power and notice that $\int_{0}^{2 π} e^{i m x} dx = 0$ 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:

$\sin^{2 n} x = \sin^{2 n - 1} x \sin x$

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

$\int u dv / dx dx = u v - \int v du / dx dx$

then use the result $\cos^{2} x = 1 - \sin^{2} x$ .