∫2,0√(4-x^2)dx怎么算

题目

∫2,0√(4-x^2)dx怎么算

答案

令x=2sint,dx=2costdt,x=2时,t= π/2,x=0时,t=0
∫(2,0)√(4-x^2)dx
=∫(π/2,0)2cost*2costdt
=2∫(π/2,0)2cos²tdt
=2∫(π/2,0)(1-cos2t)dt
=2t-sin2t│(π/2,0)
=π