分子是x的3次方乘以arccosx,分母是根号下1减x的平方求不定积分

问题描述:

分子是x的3次方乘以arccosx,分母是根号下1减x的平方求不定积分

原式=-∫{x^3arccosx/[-√(1-x^2)]}dx
  =-∫x^3arccosxd(arccosx)
  =-(1/2)∫x^3d[(arccosx)^2]
  =-(1/2)x^3(arccosx)^2+(1/2)∫(arccosx)^2d(x^3)
  =-(x^3/2)(arccosx)^2+(3/2)∫x^2(arccosx)^2dx.
令arccosx=t,则:x=cost,dx=-sintdt.
∴原式=-(x^3/2)(arccosx)^2+(3/2)∫t^2(cost)^2(-sint)dt
   =-(x^3/2)(arccosx)^2-(3/2)∫t^2[1-(sint)^2]sintdt
   =-(x^3/2)(arccosx)^2-(3/2)∫t^2sintdt+(3/2)∫t^2(sint)^3dt
   =-(x^3/2)(arccosx)^2-(3/2)∫t^2sintdt+(3/2)∫t^2[(3sint-sin3t)/4]dt
   =-(x^3/2)(arccosx)^2-(3/2)∫t^2sintdt+(9/8)∫t^2sintdt-(3/8)∫t^2sin3tdt
   =-(x^3/2)(arccosx)^2-(3/8)∫t^2sintdt-(1/8)∫t^2sin3td(3t)
   =-(x^3/2)(arccosx)^2+(3/8)∫t^2d(cost)+(1/8)∫t^2d(cos3t)
   =-(x^3/2)(arccosx)^2+(3/8)t^2cost-(3/8)∫costd(t^2)+(1/8)t^2cos3t
    -(1/8)∫cos3td(t^2)
   =-(x^3/2)(arccosx)^2+(3/8)x(arccosx)^2
    +(1/8)[cos3(arccosx)](arccosx)^2-(3/4)∫tcostdt-(1/4)∫tcos3tdt
   =(3x/8)(arccosx)^2-(x^3/2)(arccosx)^2
    +(1/8){4[cos(arccosx)]^3-3[cos(arccosx)]}(arccosx)^2
    -(3/4)∫td(sint)-(1/12)∫td(sin3t)
   =(3x/8)(arccosx)^2-(x^3/2)(arccosx)^2+(1/2)x^3(arccosx)^2
    -(3/8)x(arccosx)^2-(3/4)tsint+(3/4)∫sintdt-(1/12)tsin3t
    +(1/12)∫sin3tdt
   =-(3/4)tsint-(1/12)t[3sint-4(sint)^3]-(3/4)cost-(1/36)cos3t+C
   =-(5/6)tsint+(1/3)t(sint)^3-(3/4)x-(1/36)[4(cost)^3-3cost]+C
   =-(5/6)arccosx√[1-(cost)^2]
    +(1/3)arccosx[1-(cost)^2]√[1-(cost)^2]-(1/36)(4x^3-3x)+C
   =-(5/6)√(1-x^2)arccosx+(1/3)(1-x^2)√(1-x^2)arccosx
    -(1/9)x^3+(1/12)x+C
   =x/12-x^3/9-(1/2)√(1-x^2)arccosx-(1/3)x^2√(1-x^2)arccosx+C