Results 1 to 3 of 3

Math Help - Integrals

  1. #1
    Junior Member
    Joined
    Dec 2008
    From
    America, Ohio
    Posts
    32

    Integrals

    Hi,

    Evaluate the following integrals using integration by parts:

    a) f x sin 5x dx

    b) f x^2 e^3x dx

    * note: these functions above, are indefinate integrals

    thanks,

    moon
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Moon Hoplite View Post
    a) f x sin 5x dx
    \int x \sin 5 x dx = \int x \left( -\frac{1}{5}\cos 5x \right) ' dx = - \frac{x}{5}\cos 5x + \frac{1}{5}\int \cos 5x dx

    b) f x^2 e^3x dx
    \int x^2 \left( \frac{1}{3}e^{3x} \right) ' dx = \frac{x^2}{3}e^{3x} - \frac{2}{3} \int x e^{3x} dx

    Can you continue?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    145
    Quote Originally Posted by Moon Hoplite View Post
    Hi,

    Evaluate the following integrals using integration by parts:

    a) f x sin 5x dx

    b) f x^2 e^3x dx

    * note: these functions above, are indefinate integrals

    thanks,

    moon
    Hey!

    So, integrating by parts has this really helpful mnemonic device associated with it. ILATE - it's a priority list for what you should make your u. I - inverse trig functions L - log functions A - algebraic functions T - trig functions E - exponential functions. It's a listing of the easiest things to take the derivative of that will also simplify with each successive iteration.

    a)  \int x\ sin 5x dx

    Start by making your u = x and then your du = dx
    dv = sin(5x) and so your v = -1/5cos(5x)

     -1/5x * cos(5x) - \int -1/5 * cos(5x) dx = -1/5x * cos(5x) + 1/25 sin[5x]  + C

    b)  \int x^2\  * e^3x dx
    u = x^2
    du = 2x dx

    dv = e^3x dx
    v = (e^3x)/3

    x^2 * (e^3x)/3 - 1/3 * \int e^3x * 2x dx

    Use by parts for the integral again.

    u = 2x
    du = 2 dx

    dv = e^3x dx
    v = 1/3 * e^3x

    x^2 * (e^3x)/3 - 1/3 * ( 2x * 1/3 * e^3x - 2/3 * \int e^3x dx = ((x^2 * e^3x) - 1/3 ( (2x * e^3x)/3 - (2 *e^3x)/9 ) + C
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Contour Integrals (to Evaluate Real Integrals)
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: January 17th 2011, 10:23 PM
  2. Replies: 1
    Last Post: December 6th 2009, 08:43 PM
  3. Integrals : 2
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 24th 2009, 08:40 AM
  4. Integrals and Indefinite Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 9th 2009, 05:52 PM
  5. integrals Help please
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2008, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum