定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1
问题描述：
定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1
答
令x=√2sint
则原式=∫(π/4→π/2)√2cost*√2costdt
=∫(π/4→π/2)2cos^2(t)dt
=∫(π/4→π/2)(cos(2t)+1)dt
=sin(2t)/2|(π/4→π/2)+t|(π/4→π/2)
=0-1/2+π/4=π/4-1/2