Question

定积分的问题

根号下4-X^2是有哪个式子的导数化来得?

Answer 1

∫ √(4-x^2) dx
= x√(4-x^2) - ∫ xd√(4-x^2)
= x√(4-x^2) - ∫ -x^2/√(4-x^2) dx
= x√(4-x^2) - ∫ (4-x^2)/√(4-x^2) dx + ∫ 4/√(4-x^2) dx
= x√(4-x^2) - ∫ √(4-x^2) dx + ∫ 4/√(1-(x/2)^2) d(x/2)
= 1/2 [ x√(4-x^2) + 4arcsin(x/2) ] + C

Answer 2

arcsin(x/a)