Question

求积分∫√(1-x^2) dx

Answer 1

令x=sina
则√(1-x²)=cosa
dx=cosada

原式=∫cos²ada
=∫(1+cos2a)/2da
=1/4∫(1+cos2a)d2a
=1/4*(2a+sin2a)
【【不清楚，再问；满意， 请采纳！祝你好运开☆！！】】

Answer 2

设x=sint,dx=costdt,
t=arcsinx,
cost=√(1-x^2)
原式=∫cost*costdt
=(1/2)∫(1+cos2t)dt
=t/2+(sin2t)/4+C
=(1/2)arcsinx+2x√(1-x^2)/4+C
=(1/2)arcsinx+(x/2)√(1-x^2)+C.