定积分∫[0,a] √a^2 -x^2dx(a>0)求详细过程

2024-11-15 10:34:36
推荐回答(1个)
回答1:

∫[0,a] √宽纳a^2 -x^2dx
首先算不定积分
∫√(a^2 -x^2)dx
设x=asinu,dx=acosudu,√(a^2 -x^2)=acosu
∫码搏√(a^2 -x^2)dx
=∫acosuacosudu
=∫a^2cos^2udu
=∫慎模没a^2(1+cos2u)/2du
=a^2/4∫(1+cos2u)d(2u)
=a^2/4[2u+sin2u]