Page 1 of 2 12 LastLast
Results 1 to 15 of 22

Math Help - Integration of Rational Functions by partial fractions

  1. #1
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167

    Integration of Rational Functions by partial fractions

    I understand how to do it but I get stuck really easily. Like this for example

    \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}

    Yeah it's probably an easy one but I'm not that great with limit of integration. so I get

    x = 4sin(o)
    dx = 4cos(o)

    So here's where I am stuck. I don't know what to do now. I plug those in but I don't know what my limit of integration is supposed to be. This stuff is killing me. I'm trying really hard to understand. Please someone help me.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    12
    Quote Originally Posted by FalconPUNCH! View Post
    I understand how to do it but I get stuck really easily. Like this for example

    \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}
    Much better to substitute u^2=16-x^2.

    Quote Originally Posted by FalconPUNCH! View Post
    x = 4sin(o)
    For x=0\implies 0=4\sin\theta. For x=2\sqrt3\implies 2\sqrt3=4\sin\theta. Compute \theta in both cases.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Quote Originally Posted by Krizalid View Post
    Much better to substitute u^2=16-x^2.


    For x=0\implies 0=4\sin\theta. For x=2\sqrt3\implies 2\sqrt3=4\sin\theta. Compute \theta in both cases.
    So in order to solve for what the new limits of integration I have to set what I'm substituting like in this example sin(o) = to the old limits of integration and solve. Thanks Krizalid.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    If you must use trig sub, you have the correct subs.

    Did you sub them in and simplify?. Also, you have to change your limits of integration because you're now in terms of theta.

    x=4sin({\theta}), \;\ dx=4cos({\theta})d{\theta}

    Therefore, solve for theta using our old limits.

    2\sqrt{3}=4sin({\theta}); \;\ {\theta}=\frac{\pi}{3}

    \int\frac{64sin^{3}({\theta})}{\sqrt{16-(4sin({\theta}))^{2}}}\cdot{4cos({\theta})}d{\thet  a}

    =64\int_{0}^{\frac{\pi}{3}}sin^{3}({\theta})d{\the  ta}

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

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, FalconPUNCH!

    \int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}\,dx
    You said: . x \:= \:4\sin\theta\quad\Rightarrow\quad dx \:= \:4\cos\theta\,d\theta

    . . Substitute: .  \int\frac{64\sin^3\!\theta}{4\cos\theta}(4\cos\the  ta\,d\theta) \;=\;64\!\!\int\!\sin^3\!\theta\,d\theta



    To change the limits, recall your substitution:
    . . . x \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{4}

    When x = 0, we have: . \sin\theta \:=\:\frac{0}{4} \:=\:0\quad\Rightarrow\quad\boxed{\theta \:=\:0}

    When x = 2\sqrt{3}, we have: . \sin\theta \:=\:\frac{2\sqrt{3}}{4} \:=\:\frac{\sqrt{3}}{2}\quad\Rightarrow\quad\boxed  {\theta \:=\:\frac{\pi}{3}}

    . . And there are the new limits . . .


    The integral becomes: . 64\!\int^{\frac{\pi}{3}}_0\!\!\sin^3\!\theta\,d\th  eta


    Got it?

    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Quote Originally Posted by galactus View Post
    If you must use trig sub, you have the correct subs.

    Did you sub them in and simplify?. Also, you have to change your limits of integration because you're now in terms of theta.

    x=4sin({\theta}), \;\ dx=4cos({\theta})d{\theta}

    Therefore, solve for theta using our old limits.

    2\sqrt{3}=4sin({\theta}); \;\ {\theta}=\frac{\pi}{3}

    \int\frac{64sin^{3}({\theta})}{\sqrt{16-(4sin({\theta}))^{2}}}\cdot{4cos({\theta})}d{\thet  a}

    =64\int_{0}^{\frac{\pi}{3}}sin^{3}({\theta})d{\the  ta}

    Can you finish?.
    I got up to the last part the only thing I was having trouble with was the limits of integration.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Quote Originally Posted by FalconPUNCH! View Post
    I got up to the last part the only thing I was having trouble with was the limits of integration.

    Quote Originally Posted by Soroban View Post
    Hello, FalconPUNCH!

    You said: . x \:= \:4\sin\theta\quad\Rightarrow\quad dx \:= \:4\cos\theta\,d\theta

    . . Substitute: .  \int\frac{64\sin^3\!\theta}{4\cos\theta}(4\cos\the  ta\,d\theta) \;=\;64\!\!\int\!\sin^3\!\theta\,d\theta



    To change the limits, recall your substitution:
    . . . x \:=\:4\sin\theta \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{4}

    When x = 0, we have: . \sin\theta \:=\:\frac{0}{4} \:=\:0\quad\Rightarrow\quad\boxed{\theta \:=\:0}

    When x = 2\sqrt{3}, we have: . \sin\theta \:=\:\frac{2\sqrt{3}}{4} \:=\:\frac{\sqrt{3}}{2}\quad\Rightarrow\quad\boxed  {\theta \:=\:\frac{\pi}{3}}

    . . And there are the new limits . . .


    The integral becomes: . 64\!\int^{\frac{\pi}{3}}_0\!\!\sin^3\!\theta\,d\th  eta


    Got it?

    yeah it's pretty clear now. Thanks guys
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    So the answer I get for this is \frac{24\sqrt{3} - 128}{3}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    No, I am sorry to say, that is incorrect.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Dec 2007
    Posts
    131
    You really shouldn't let the limits of integration stop you from solving the problem. If you are having trouble finding the new limits, simply forget about them until you have to evaluate your u function. Then just "un-substitute" your original function back in place of u and use the original limits.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    I'm really stressing out over this. I can't even solve a single integral because of this I'm trying to do the ones ahead but they're even more annoying. I might do what you said, xifentoozlerix but I get confused when I reach the end.

    Here's what I'm doing:

    64\int_{0}^{\frac{pi}{3}} sin^{2}0 sin(0) d0

     u = cos(o), du = -sin(o)do


    64\int_{0}^{\frac{pi}{3}} (1-cos^{2}x) sin(0) d0


    64\int_{0}^{\frac{pi}{3}} (1 - u^2) -du

     -64( u - \frac{1}{3}u^{3})

     -64( cos(o) - \frac{1}{3}cos(o)^{3})

    I then plugged in pi/3 and 0 and the answer I posted before is what I got. I'm not sure where I messed up.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Perhaps try the advice that was given about waiting 'til the end.

    You can always use the reduction formula for sine. Save you all the grunt work.

    \int{sin^{n}(u)}du=\frac{-1}{n}sin^{n-1}(u)cos(u)+\frac{n-1}{n}\int{sin^{n-2}}du
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Okay I haven't used to reduction formula a lot so I'm not sure if I did it correctly

     64( \frac{-1}{3}sin^{2}(u)cos(u) + \frac{2}{3} \int sin du)

     64( \frac{-1}{3}sin^{2}(\frac{pi}{3})cos(\frac{pi}{3}) - sin^{2}(0)cos(0) + \frac{2}{3} ( \frac{1}{2} - \frac{pi}{2} )

     64(( (\frac{-1}{3}) (\frac{3}{4})(\frac{1}{2}) - \frac{pi}{2}) + \frac{2}{3} (\frac{1}{2} - \frac{pi}{2})

    I then simplify it to

    64( \frac{-8 - 23pi}{24})

    not sure if it's right...
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by FalconPUNCH! View Post
    Okay I haven't used to reduction formula a lot so I'm not sure if I did it correctly

     64( \frac{-1}{3}sin^{2}(u)cos(u) + \frac{2}{3} \int sin du)

     64( \frac{-1}{3}sin^{2}(\frac{pi}{3})cos(\frac{pi}{3}) - sin^{2}(0)cos(0) + \frac{2}{3} ( \frac{1}{2} - \frac{pi}{2} )

     64(( (\frac{-1}{3}) (\frac{3}{4})(\frac{1}{2}) - \frac{pi}{2}) + \frac{2}{3} (\frac{1}{2} - \frac{pi}{2})

    I then simplify it to

    64( \frac{-8 - 23pi}{24})

    not sure if it's right...
    I = 64 \int_{0}^{\pi/3} \sin^3 \theta \, d\theta = 64 \int_{0}^{\pi/3} \sin^2 \theta \, \sin \theta \, d\theta = 64 \int_{0}^{\pi/3} (1 - \cos^2 \theta) \sin \theta \, d\theta.

    Substitute u = \cos \theta. Then:

    I = -64 \int_{1}^{1/2} 1 - u^2 \, du = ......

    I get I = \frac{40}{3}.
    Last edited by mr fantastic; February 21st 2008 at 05:00 PM.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Member FalconPUNCH!'s Avatar
    Joined
    Sep 2007
    From
    Daly City, CA
    Posts
    167
    Quote Originally Posted by mr fantastic View Post
    I = 64 \int_{0}^{\pi/3} \sin^3 \theta \, d\theta = 64 \int_{0}^{\pi/3} \sin^2 \theta \, \sin \theta \, d\theta = 64 \int_{0}^{\pi/3} (1 - \cos^2 \theta) \sin \theta \, d\theta.

    Substitute u = \cos \theta. Then:

    I = -64 \int_{1}^{1/2} 1 - u^2 \, du = ......
    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?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Partial Fractions/ Integration
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 13th 2010, 03:53 PM
  2. Replies: 5
    Last Post: August 5th 2010, 03:30 AM
  3. Replies: 4
    Last Post: October 2nd 2008, 07:20 PM
  4. Replies: 7
    Last Post: June 22nd 2008, 12:59 PM
  5. Replies: 7
    Last Post: June 19th 2008, 10:33 AM

Search Tags


/mathhelpforum @mathhelpforum