Question

(1-x)/根号下(9-x^2)的不定积分

Answer 1

令x=3sina
dx=3cosada
√(9-x²)=3cosa
a=arcsin(x/3)

sina=x/3
则cosa=√(1-x²/9)=√(9-x²)/3

所以原式=∫(1-3sina)/3cosa*3cosada
=∫(1-3sina)da
=a+3cosa+C
=arcsin(x/3)+√(9-x²)/3+C