Question

根号（a^2-x^2)dx的形式怎么求?

Answer 1

令x=asint,t=arcsinx/a
原式=∫√[a^2(1-sint·sint)]acostdt
=a^2∫(cost)^2dt
=a^2∫1/2+1/2cost/2dt
=a^2t/2+∫cott/2dt/2
=a^2t/2+sint/2+c
=(a^2arcsinx/a)/2+sin[(arcsinx/a)/2]+c