Question

求不定积分题

Answer 1

令 x = cosu
I = ∫(cosu)^3 u(-sinu)du/sinu = -∫(cosu)^3 udu
= -∫u[1-(sinu)^2]dsinu = -∫ud[sinu-(1/3)(sinu)^3]
= -u[sinu-(1/3)(sinu)^3] + ∫[sinu-(1/3)(sinu)^3]du
= -u[sinu-(1/3)(sinu)^3] - cosu + (1/3)∫[1-(cosu)^2]dcosu
= -u[sinu-(1/3)(sinu)^3] - cosu + (1/3)[cosu-(1/3)(cosu)^3] + C
= -u[sinu-(1/3)(sinu)^3] - (2/3)cosu - (1/9)(cosu)^3 + C
= -(1/3)(2+x^2)√(1-x^2)arccosx - (2/3)x - (1/9)x^3 + C

Answer 2

∫ x^3.arccosx/√(1-x^2 ) dx
= -∫ x^2.arccosx d√(1-x^2 )
=-x^2.√(1-x^2 ).arccosx +∫[ 2x.√(1-x^2 ).arccosx - x^2 ] dx
=-x^2.√(1-x^2 ).arccosx - (1/3)x^3 +2∫x.√(1-x^2 ).arccosx dx
=-x^2.√(1-x^2 ).arccosx - (1/3)x^3 -(2/3)∫arccosx d(1-x^2 )^(3/2)
=-x^2.√(1-x^2 ).arccosx - (1/3)x^3 -(2/3)(1-x^2 )^(3/2).arccosx
- (2/3)∫√(1-x^2 ) dx
=-x^2.√(1-x^2 ).arccosx - (1/3)x^3 -(2/3)(1-x^2 )^(3/2).arccosx
- (1/3)[ arcsinx+ x/(1+x^2) ] + C
//
let
x=sinu
dx =cosu du
∫√(1-x^2 ) dx
=∫ (cosu)^2 du
=(1/2)∫ (1+cos2u) du
=(1/2)[u+(1/2)sin2u] +C'
=(1/2)[ arcsinx+ x/(1+x^2) ]+ C'

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