Results 1 to 8 of 8

Math Help - Integration By Parts

  1. #1
    Newbie
    Joined
    Apr 2007
    Posts
    14

    Integration By Parts

    I'm having some trouble with my math homework again...

    Use integration by parts to find the indefinite integral:
    1) INT 1/(x*(lnx)^3) dx
    2) INT x/sqroot(2+3x) dx
    3) INT (x+sin3x)^2 dx
    4) INT sin x^(1/3) dx
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by cupcakelova87 View Post
    Use integration by parts to find the indefinite integral:
    1) INT 1/(x*(lnx)^3) dx
    Integration by parts for this? Substitution would work better here.

    Let u = lnx <--> du = 1/x dx --> dx = x du

    INT 1/(x*u^3)*x du
    = INT 1/u^3 du
    = -1/(2u^2) + C
    = -1/(2(lnx)^2) + C
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by cupcakelova87 View Post
    Use integration by parts to find the indefinite integral:

    2) INT x/sqroot(2+3x) dx
    Let u = x <--> du = dx
    Let dv = 1/sqrt(2 + 3x) dx <--> v = (2/3)sqrt(2 + 3x)

    => (2/3)x*sqrt(2 + 3x) - INT (2/3)sqrt(2 + 3x) dx
    = (2/3)x*sqrt(2 + 3x) - (4/27)sqrt(2 + 3x)^3 + C
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2007
    Posts
    14
    Thanks ecmathgeek! Quick question...how did you get V for problem #2?

    "Let dv = 1/sqrt(2 + 3x) dx <--> v = (2/3)sqrt(2 + 3x)"
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by cupcakelova87 View Post
    Use integration by parts to find the indefinite integral:

    3) INT (x+sin3x)^2 dx
    First, I would simplify the squared term:
    INT x^2 + 2xsin3x + sin^2(3x) dx

    Now this integration has three parts: x^2 (which is easy), 2xsin3x (which needs by-parts), sin^2(3x) (which needs trig identity).
    INT x^2 dx + 2*INT xsin3x dx + INT sin^2(3x) dx

    For INT x^2 dx:
    INT x^2 dx
    = 1/3*x^3

    For INT 2xsin3x dx:
    Let u = 2x <--> du = 2dx
    Let dv = sin3x dx <--> v = -1/3*cos3x
    => -(2/3)xcos3x - 2/3*INT cos3x dx
    = -(2/3)xcos3x - (2/9)sin3x

    For INT sin^2(3x) dx:
    Let sin^2(3x) = 1/2(1 - cos6x) <-- this is the half-angle formula for sin(3x) after squaring both sides.
    1/2*INT 1 - cos6x dx
    = 1/2*(x - 1/6*sin6x)
    = (1/2)x - (1/12)sin6x

    Recombining these integrations, we get:
    (1/3)x^3 - (2/3)xcos3x - (2/9)sin3x + (1/2)x - (1/12)sin6x + C
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by cupcakelova87 View Post
    Thanks ecmathgeek! Quick question...how did you get V for problem #2?

    "Let dv = 1/sqrt(2 + 3x) dx <--> v = (2/3)sqrt(2 + 3x)"
    Good question. I skipped that step, but I'll explain it.

    We have:
    dv = 1/sqrt(2 + 3x) dx

    To get v, we need to integrate both sides:
    INT dv = INT 1/sqrt(2 + 3x) dx

    The integration of dv is easy. The integration of 1/sqrt(2 + 3x) requires an additional substitution.
    Let n = 2 + 3x <--> dn = 3 dx --> dx = 1/3 dn
    1/3*INT 1/sqrt(n) dn
    = 1/3*INT (n)^(-1/2) dn
    = 1/3*(2/1)*n^(1/2)
    = (2/3)sqrt(2 + 3x)

    Therefore:
    v = (2/3)sqrt(2 + 3x)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Anytime I show something like:
    Let dv = f(x) dx <--> v = F(x)

    Where F(x) is the antiderivative of f(x), I recommend you try to figure out how I went from f(x) dx to F(x). Try doing the integration yourself. It should improve your ability to do integrations.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member ecMathGeek's Avatar
    Joined
    Mar 2007
    Posts
    436
    Quote Originally Posted by cupcakelova87 View Post
    Use integration by parts to find the indefinite integral:

    4) INT sin x^(1/3) dx
    Let u = sinx^(1/3) <--> du = 1/3*x^(-2/3)*sinx^(1/3) dx
    Let dv = dx <--> v = x

    => xsinx^(1/3) - (1/3)*INT x^(-2/3)sinx^(1/3) dx

    Let w = x^(1/3) <--> dw = (1/3)x^(-2/3) dx --> dx = 3*x^(2/3) dw

    => xsinx^(1/3) - INT sinw dw
    = xsinx^(1/3) + cosw + C
    = xsinx^(1/3) + cosx^(1/3) + C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 03:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 04:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 04:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 07:55 AM

Search Tags


/mathhelpforum @mathhelpforum