Question

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？

Answer 1

∫[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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