(sinx)^2n在0到π/2积分怎么算
问题描述：
(sinx)^2n在0到π/2积分怎么算
答
I(2n)=∫(0->π/2) (sinx)^(2n) dx
=-∫(0->π/2) (sinx)^(2n-1) dcosx
= [-cosx.(sinx)^(2n-1)](0->π/2) +(2n-1)∫(0->π/2) (cosx)^2.(sinx)^(2n-2) dx
= (2n-1)∫(0->π/2)[ 1-(sinx)^2].(sinx)^(2n-2) dx
2nI(2n) = (2n-1)I(2n-2)
I(2n)= [(2n-1)/(2n)]I(2n-2)
= { (2n-1)(2n-3).1/[(2n(2n-2)(2n-4).2) ] }. I0
= { (2n-1)(2n-3).1/[(2n(2n-2)(2n-4).2) ] }∫(0->π/2) dx
= { (2n-1)(2n-3).1/[(2n(2n-2)(2n-4).2) ] } (π/2)