Page 2 of 2 FirstFirst 12
Results 16 to 22 of 22

Math Help - Integration of Rational Functions by partial fractions

  1. #16
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    With trig sub it whittlles down nicely.

    Let x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}

    Make the subs and you get:

    \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Quote Originally Posted by galactus View Post
    With trig sub it whittlles down nicely.

    Let x=sec({\theta}), \;\ dx=sec({\theta})tan({\theta})d{\theta}

    Make the subs and you get:

    \int_{0}^{\frac{\pi}{3}}tan^{2}({\theta})d{\theta}
    yeah I did it that way, but I got stuck on the limits of integration I don't know what  1 = sec({\theta}) is. Is there a quick way to figure out what they are?
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by FalconPUNCH! View Post
    yeah I did it that way, but I got stuck on the limits of integration I don't know what  1 = sec({\theta}) is. Is there a quick way to figure out what they are?
    sec(\theta) = 1

    \frac{1}{cos(\theta)} = 1

    cos(\theta) = 1

    \theta = 0

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by FalconPUNCH! View Post
    I get \frac{40}{3}

    hmmm...seem to be getting the limits of integration wrong which is happening in my homework a lot.

    for this one \int_{1}^{2} \frac{\sqrt{x^{2} - 1}}{x}dx is it best to use u substitution from the start?
    Alternatively, make the substitution u^2 = x^2 - 1.

    Then the integral becomes

    I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du.

    Details can be given if needed .....
    Follow Math Help Forum on Facebook and Google+

  5. #20
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by mr fantastic View Post
    Alternatively, make the substitution u^2 = x^2 - 1.

    Then the integral becomes

    I = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du.

    Details can be given if needed .....
    u^2 = x^2 - 1 \Rightarrow 2u \frac{du}{dx} = 2x \Rightarrow dx = \frac{u}{x} \, du

    x = 1 \Rightarrow u^2 = 0 \Rightarrow u = 0. x = 2 \Rightarrow u^2 = 3 \Rightarrow u = \sqrt{3} where the positive root is used.

    Then:

    I = \int_{0}^{\sqrt{3}} \frac{u}{x} \frac{u}{x} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{x^2} \, du = \int_{0}^{\sqrt{3}} \frac{u^2}{u^2 + 1} \, du


     = \int_{0}^{\sqrt{3}} \frac{(u^2 + 1) - 1}{u^2 + 1} \, du = \int_{0}^{\sqrt{3}} 1 - \frac{1}{u^2 + 1} \, du,


    which integrates easily to give I = \sqrt{3} - \frac{\pi}{3}.
    Follow Math Help Forum on Facebook and Google+

  6. #21
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these.
    Follow Math Help Forum on Facebook and Google+

  7. #22
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by FalconPUNCH! View Post
    Thank you guys for helping me out. I'm going to use this thread as a reference so I won't forget how to do these.
    Good thinking, 99
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. Partial Fractions/ Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 13th 2010, 04:53 PM
  2. Replies: 5
    Last Post: August 5th 2010, 04:30 AM
  3. Replies: 4
    Last Post: October 2nd 2008, 08:20 PM
  4. Replies: 7
    Last Post: June 22nd 2008, 01:59 PM
  5. Replies: 7
    Last Post: June 19th 2008, 11:33 AM

Search Tags


/mathhelpforum @mathhelpforum