Question

高数问题求大神解答？

29题

Answer 1

let
x=sinu
dx =cosu du
∫ dx/[1+√(1-x^2)]
=∫ cosu/(1+cosu) du
=∫ [1 - 1/(1+cosu)] du
=u -∫ [ 1/(1+cosu)] du
=u -∫ (1-cosu)/(sinu)^2 du
=u -∫ [ (cscu)^2 -cosu/(sinu)^2 ]du
=u -[ -cotu + 1/sinu ] +C
=u + cotu - 1/sinu +C
=arcsinx + √(1-x^2)/x - 1/x +C

Answer 2

如图片