Question

求(10∧(2arccosx))/√(1-x²)dx的不定积分

Answer 1

∫10^2arccosx/√1-x²dx
=-1/2 ∫10^2arccosxd2arccosx
= -10^2arccosx/2ln10 + C

Answer 2

∫ arccos x dx
=xarccosx-∫xdarccosx
=xarccosx+∫x/√(1-x²)dx
=xarccosx+∫d(1-x²)/2√(1-x²)
=xarccosx+√(1-x²)+C