这个怎么求原函数啊？

Question

如题所述

Answer 1

=1/2∫√((1-r²)/(1+r²))dr²
三角换元脱根号令r²=sinu
=1/2∫(0.π/2)√((1-sinu)(1+sinu))/(1+sinu)dsinu
=1/2∫cos²u/(1+sinu)du
=1/2∫(1-sinu)du
=(u+cosu)/2
=(π-2)/4本回答被网友采纳