Counter
Mathematics Dictionary
Dr. K. G. Shih
Integral by part
Subjects

Answers


Q01. Integral by part

Definition
  • It is corresponding the product rule in derivative
  • Product rule : d/dx(F(x)*G(x)) = F'(x)*G(x) + F(x)*G'(x)
  • Hence d(F(x)*G(x)) = (F'(x)*G(x) + F(x)*G'(x))dx
  • Hence F(x)*G(x) = ∫(F'(x)*G(x) + F(x)*G'(x))dx
  • Hence ∫F'(x)*G(x)dx = F(x)*G(x) - ∫F(x)*G'(x)dx

Go to Begin

Q02. Find ∫ ln(x)dx

  • There is no corresponding derivative, we try integral by part
  • Let dv = dx and v = x
  • Let u = ln(x) and du = (1/x)dx
  • Integral = u*v - ∫ vdu
  • Integral = x*ln(x) - ∫ x*(1/x)dx
  • Integral = x*ln(x) - x
Verufy
  • d/dx(x*ln(x) - x) = (ln(x) + x*(1/x) - 1) = ln(x)

Go to Begin

Q03. Find ∫ x*exp(x)dx

  • There is no corresponding derivative, we try integral by part
  • Let dv = exp(x)dx and v = exp(x)
  • Let u = x and du = dx
  • Integral = u*v - ∫ vdu
  • Integral = x*exp(x) - ∫ exp(x)dx
  • Integral = x*exp(x) - exp(x)
Verufy
  • d/dx(x*exp(x) - exp(x)) = (exp(x) + x*exp(x) - exp(x)) = x*exp(x)

Go to Begin

Q04. Find ∫ x*sin(x)dx

  • There is no corresponding derivative, we try integral by part
  • Let dv = sin(x)dx and v = -cos(x)
  • Let u = x and du = dx
  • Integral = u*v - ∫ vdu
  • Integral = -x*cos(x) + ∫ cos(x)dx
  • Hence ∫ x*sin(x)dx = -x*cos(x) + sin(x)
Verufy
  • d/dx(sin(x) - x*cos(x))/2 = (cos(x) - cos(x) + x*sin(x)) = x*sin(x)

Go to Begin

Q05. Find ∫ exp(x)*sin(x)dx
  • There is no corresponding derivative, we try integral by part
  • Let dv = exp(x)dx and v = exp(x)
  • Let u = sin(x) and du = cos(x)dx
  • Integral = u*v - ∫ vdu
  • Integral = exp(x)*sin(x) + ∫ exp(x)*cos(x)dx
  • What is ∫ exp(x)*cos(x)dx ?
  • Integral by part again
    • Let dv = exp(x)dx and v = exp(x)
    • Let u = cos(x) and du = -sin(x)dx
    • Integral = u*v - ∫ vdu
    • Integral = exp(x)*cos(x) - ∫ exp(x)*sin(x)dx
  • ∫ x*sin(x)dx
  • = exp(x)*sin(x) + ∫ exp(x)*cos(x)dx
  • =exp(x)*sin(x) + exp(x)*cos(x) - ∫ exp(x)*sin(x)dx
  • 2* ∫ exp(x)*sin(x)dx = exp(x)*sin(x) - exp(x)*cos(x)
  • Hence ∫ x*sin(x)dx = (exp(x)*sin(x) - exp(x)*cos(x))/2
Verify
  • d/dx(exp(x)*sin(x) - exp(x)*cos(x))/2
  • = (exp(x)*sin(x) + sxp(x)*cos(x) - exp(x)*cos(x) + exp(x)*sin(x))/2
  • = exp(x)*sin(x)
    Go to Begin

    Q06. Find ∫(x^2)*exp(x)

  • There is no corresponding derivative, we try integral by part
  • Let dv = exp(x)dx and v = exp(x)
  • Let u = x^2 and du = 2*xdx
  • Integral = u*v - ∫ vdu
  • Integral = (x^2)exp(x) - ∫ 2*x*exp(x)dx
  • What is ∫ 2*x*exp(x)dx ?
  • Integral by part again
    • Let dv = exp(x)dx and v = exp(x)
    • Let u = 2x and du = 2dx
    • Integral = u*v - ∫ vdu
    • Integral = 2*x*exp(x) - ∫ exp(x)dx = exp(x)
  • ∫ (x^2)*exp(x)dx
  • = (x^2)*exp(x) - ∫ x*exp(x)dx
  • = (x^2)*exp(x) - 2*x*exp(x) + 2*exp(x) Verify
    • d/dx((x^2)*exp(x) - 2*x*exp(x) + 2*exp(x))
    • =

    Go to Begin

    Q07. ∫(1/Sqr(1 - x^2))dx = arcsin(x)

    Use anti-derivative : d/dx(arcsin(x)) = 1/Sqr(1 - x^2)
    • d(arcsin(x)) = (1/Sqr(1 - x^2))dx
    • Hence integral of (1/Sqr(1 - x^2))dx = arcsin(x)
    Second method
    • Let x = sin(U) then dx = cos(U)dU and U = arcsin(x)
    • Hence (1/Sqr(1 - x^2)*dx = (1/Sqr(1 - sin(U)^2))*(cos(U))dU
    • Since 1 - sin(U)^2 = cos(U)^2
    • Hence (1/Sqr(1 - x^2)*dx = dU
    • Hence ∫(1/Sqr(1 - x^2))dx = ∫dU = U = arcsin(x) + C

    Go to Begin

    Q08. ∫(-1/Sqr(1-x^2))dx = arccos(x)

    Use anti-derivative : d/dx(arccos(x)) = -1/Sqr(1 - x^2)
    • d(arccos(x)) = (-1/Sqr(1 - x^2))dx
    • Hence integral of (-1/Sqr(1 - x^2))dx = arccos(x)
    Second method
    • Let x = cos(U) then dx = -sin(U)dU and U = arccos(x)
    • Hence (-1/Sqr(1 - x^2)*dx = (-1/Sqr(1 - cos(U)^2))*(-sin(U))dU
    • Since 1 - cos(U)^2 = sin(U)^2
    • Hence (-1/Sqr(1 - x^2)*dx = dU
    • Hence ∫(-1/Sqr(1 - x^2))dx = ∫dU = U = arccos(x) + C

    Go to Begin

    Q09. ∫(1/(1 + x^2))dx = arctan(x)

    Use derivative d/dx(arctan(x)) = 1/(1 + x^2)
    • d(arctan(x)) = (1/(1 + x^2))dx
    • Hence integral of (1/(1+x^2))dx = arctan(x)
    Second method
    • Let x = tan(U) then dx = sec(U)^2)dU and U = arctan(x)
    • Hence (1/(1 + x^2)*dx = (1/(1 + tan(U)^2))*(sec(U)^2)dU
    • Since 1 + tan(U)^2 = sec(U)^2
    • Hence (1/(1 + x^2)*dx = dU
    • Hence ∫(1/(1 + x^2))dx = ∫dU = U = arctan(x) + C

    Go to Begin

    Q10. ∫(csc(x)^2)dx = ?

    There is anti-derivative.
    • Since d/dx(cot(x)) = -csc(x)^2
    • Hence d(cot(x)) = (-csc(2*x)^2)dx
    • ∫(-csc(x)^2)dx = cot(x) + C
    Verify
    • d/dx(cot(x)) = -csc(x)^2

    Go to Begin

    Q11. ∫(sec(x)^2)dx = ?

    There is anti-derivative.
    • Since d/dx(tan(x)) = sec(x)^2
    • Hence d(tan(x)) = (sec(2*x)^2)dx
    • ∫(sec(x)^2)dx = tan(x) + C
    Verify
    • d/dx(tan(x)) = sec(x)^2

    Go to Begin

    Q12. ∫(-1/(1 + x^2))dx = arccot(x)

    Use derivative d/dx(arccot(x)) = -1/(1 + x^2)
    • d(arccot(x)) = (-1/(1 + x^2))dx
    • Hence integral of (-1/(1+x^2))dx = arccot(x)
    Second method
    • Let x = cot(U) then dx = (-csc(U)^2)dU and U = arccot(x)
    • Hence (-1/(1 + x^2)*dx = (1/(1 + cot(U)^2))*(-csc(U)^2)dU
    • Since 1 + tan(U)^2 = csc(U)^2
    • Hence (-1/(1 + x^2)*dx = dU
    • Hence ∫(-1/(1 + x^2))dx = ∫dU = U = arccot(x) + C

    Go to Begin

    Q13. New
    Go to Begin

    • Show Room of MD2002 Contact Dr. Shih Math Examples Room
    Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
    1