定积分∫[0,a] √a^2 -x^2dx(a>0)求详细过程

Question

如题所述

Answer 1

∫[0,a] √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
∫√（a^2 -x^2）dx
=∫acosuacosudu
=∫a^2cos^2udu
=∫a^2(1+cos2u)/2du
=a^2/4∫(1+cos2u)d(2u)
=a^2/4[2u+sin2u]