题目

定积分问题：下限0上限π ∫ (sinx)的m次方 dx为什么等于2 ∫下限0上限π/2 (sinx)次方
如何证明?

提问时间：2020-08-03

答案

∫[0→π] (sinx)^m dx
=∫[0→π/2] (sinx)^m dx + ∫[π/2→π] (sinx)^m dx
后一部分做变量替换,令x=π-u,则dx=-du,u：π/2→0
=∫[0→π/2] (sinx)^m dx - ∫[π/2→0] (sin(π-u))^m du
=∫[0→π/2] (sinx)^m dx - ∫[π/2→0] (sinu)^m du
=∫[0→π/2] (sinx)^m dx + ∫[0→π/2] (sinu)^m du
积分变量可随便换字母
=2∫[0→π/2] (sinx)^m dx
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