求证明积分如下，谢谢，请尽量详细

Question

如题所述

Answer 1

最后一步，因为(cosx)^10以2π为周期，用到周期函数的定积分性质：周期为T的函数，在任何长度为T的区间上积分，值相等。

Answer 2

∫(0->2π) (sinx)^10 dx
=-∫(0->2π) (sinx)^9 dcosx
=-[(sinx)^9 .cosx] |(0->2π) +9∫(0->2π) (sinx)^8 .(cosx)^2 dx
=9∫(0->2π) (sinx)^8 .( 1- (sinx)^2) dx
10∫(0->2π) (sinx)^10 dx = 9∫(0->2π) (sinx)^8 dx
∫(0->2π) (sinx)^10 dx
= (9/10)∫(0->2π) (sinx)^8 dx
=(9/10) (7/8)∫(0->2π) (sinx)^6 dx
=...
=(9/10) (7/8)(5/6)(3/4)(1/2) ∫(0->2π) dx
=(9/10) (7/8)(5/6)(3/4)(1/2)(2π)
Similarly

∫(0->2π) (cosx)^10 dx
=∫(0->2π) (cosx)^9 dsinx
=[(cosx)^9 .sinx]|(0->2π) +9∫(0->2π) (cosx)^8.(sinx)^2 dx
10∫(0->2π) (cosx)^10 dx= 9∫(0->2π) (cosx)^8 dx
∫(0->2π) (cosx)^10 dx
= (9/10)∫(0->2π) (cosx)^8 dx
=(9/10) (7/8)(5/6)(3/4)(1/2) ∫(0->2π) dx
=(9/10) (7/8)(5/6)(3/4)(1/2)(2π)
= ∫(0->2π) (sinx)^10 dx