证明∫(sinx)^ndx=∫(cosx)^ndx,两个的积分上限都为2π,下限都为0

Question

证明∫(sinx)^ndx=∫(cosx)^ndx,两个的积分上限都为2π,下限都为0

Answer 1

令t=π/2-x,dx=-dt
当x=0时,t=π/2,当x=2π时,t=-3π/2
故
∫[0,2π](cosx)^ndx
=∫[0,2π][sin(π/2-x)]^ndx
=-∫[0,2π][sin(π/2-x)]^nd(π/2-x)
=∫[π/2,-3π/2] (sint)^ndt
由于函数sinx的周期是2π
故上式等于∫[0,2π](sinx)^ndx
即∫(sinx)^ndx=∫(cosx)^ndx