Question

求这道题的不定积分：∫ arccosxdx

Answer 1

∫ arccosx dx

= xarccosx - ∫ x d(arccosx)

= xarccosx - ∫ x * [- 1/√(1 - x²)] dx,分部积分法

= xarccosx - (1/2)∫ 1/√(1 - x²) d(1 - x²)

= xarccosx - (1/2)(2√(1 - x²)) + C

= xarccosx - √(1 - x²) + C