sinx2020次方的定积分0+π/2？

Question

如题所述

Answer 1

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) (sinx)^(2n-2).(cosx)^2 dx
=0 +(2n-1)∫(0->π/2) (sinx)^(2n-2).[1-(sinx)^2] dx
2nI(2n)=(2n-1)I(2n-2)
I(2n)=[(2n-1)/(2n)]I(2n-2)
∫(0->π/2) (sinx)^(2020) dx
=I(2020)
=(2019/2020).I(2018)
=(2019/2020).(2017/2018).I(2016)
=(2019/2020).(2017/2018)...(1/2).I0
=(2019/2020).(2017/2018)...(1/2).∫(0->π/2) dx
=(2019/2020).(2017/2018)...(1/2)(π/2)