Question

1/（1+√（1-x^2））的不定积分

1/（1+√（1-x^2））的不定积分，谢谢

Answer 1

令x=sinz，dx=coszdz
∫dx/[x+√(1-x²)]
=∫cosz/(sinz+cosz)*dz
=(1/2)∫[(sinz+cosz)+(-sinz+cosz)]/(sinz+cosz)*dz
=(1/2)∫dz + (1/2)∫d(cosz+sinz)/(sinz+cosz)*dz
=(1/2)z + (1/2)ln|sinz+cosz| + C
=(1/2)(arcsinz) + (1/2)ln|x+√(1-x²)| + C

追问

被积函数分母是1+√（1-x∧2）不是x+....

追答

∫dx/(1+√(1-x^2))
x=sinu dx=cosudu √(1-x^2)=cosu
tan(u/2)=sinu/(1+cosu)=x/(1+√(1-x^2))
=∫cosudu/(1+cosu)
=∫[1-1/(1+cosu)]du
=u-∫du/(1+cosu)
=u-∫d(u/2)/(cos(u/2))^2
=u-tan(u/2)+C
=arcsinx - x/(1+√(1-x^2)) +C