Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - [SOLVED] Integration by parts, evil arctan

  1. #1
    Junior Member
    Joined
    Jun 2008
    Posts
    68

    [SOLVED] Integration by parts, evil arctan

    <br />
\int x~arctan(7x)~dx<br />

    u= arctan(7x)

    du= \frac {1}{1+49x^2}

    dv= x~dx

    v= \frac {x^2}{2}

    This is where I am at so far:
    <br />
x^2arctan(7x) - \frac {1}{2}\int \frac {x^2}{1+49x^2}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Use polynomial long division for the second integral. You will get 2 simpler integrals.
    Last edited by Chop Suey; September 4th 2008 at 07:10 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Or, if you want to do things quickly, pull 49 as a factor:
    \frac{x^2}{49(\frac{1}{49} + x^2)}


    Can you see how to simplify this quickly without resorting to long division? Hint: Add \frac{1}{49} and subtract \frac{1}{49} in the numerator.
    Last edited by Chop Suey; September 4th 2008 at 10:18 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    We have only talked about the long division method, but I need a little help with the 2 other integrals. What will the other two integrals be?

    So I see that after long division, I get 1/49, but unfortunately I don't really know what to do next.

    <br />
\int \frac {1}{49} (1- \frac {x^2}{1+49x^2})<br />

    Is this correct? What is next?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Jun 2008
    Posts
    792
    Quote Originally Posted by redman223 View Post
    <br />
\int \frac {1}{49} (1- \frac {\color{red}x^2}{1+49x^2})<br />
    That x^2 shouldn't be there. It's supposed to be:

    \frac{1}{1+49x^2}

    Once you figure that out, notice that:

    \frac{1}{1+49x^2} = \frac{1}{1+(7x)^2}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    <br />
\int \frac {1}{49} (1- \frac {1}{1+49x^2})<br />

    So what happens to the first part of the equation? With regards to the 1/49 and the 1-?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Jun 2008
    Posts
    792
    \int \frac {1}{49} (1- \frac {1}{1+49x^2}) can be split into:

    = \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+49x^2}

    = \int \frac {1}{49} - \frac{1}{49} \int \frac {1}{1+(7x)^2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    807
    Thanks
    27
    it's \int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx okay? So you can split it into two:

    \int \frac 1 {49}\left({1 - \frac {1}{1 + 49x}}\right) dx = \int \frac 1 {49}\left({1}\right) dx  - \int \frac 1 {49}\left({\frac {1}{1 + 49x}}\right) dx
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    Thanks for clarifying that up for me. Now I can see a solution using u substitution.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    Well I thought I did, but I guess not. I am stuck once again.

    <br />
-\frac {1}{2}\int \frac {1}{49}~dx = -\frac {1}{2}(\frac {1}{49}x)<br />

    If thats the case, then
    <br />
-\frac {1}{2}(\frac {1}{49}x - \frac {1}{49}\int \frac {1}{1+49x^2}~dx)<br />

    u= 7x
    du= 7

    So in order to get the 7 in there I have to multiply 1/49 by 1/7 in order to get 1/343 right? Then would the answer just be:

    <br />
-\frac {1}{2}(\frac {arctan(7x)}{343})<br />

    for the last part of the equation?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Jun 2008
    Posts
    792
    I just noticed that you forgot to multiply by the constant 7.

    u = \arctan{7x}

    du = \frac{7}{1+49x^2} dx
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Jun 2008
    Posts
    68
    Isn't that why you pull a 1/7 out in front? Which means you would multiply 1/49 and 1/7 to get 1/343?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by redman223 View Post
    Isn't that why you pull a 1/7 out in front? Which means you would multiply 1/49 and 1/7 to get 1/343?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    you could also use a trig sub, for example say tan(u)=7x
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Banned
    Joined
    Aug 2008
    Posts
    530
    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}<br />
{2}\int {\frac{{x^2 }}<br />
{{1 + 49x^2 }}} dx \hfill \\

    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}<br />
{{2 \times 49}}\int {\frac{{49x^2 }}<br />
{{1 + 49x^2 }}} dx \hfill \\

    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}<br />
{{98}}\int {\frac{{1 + 49x^2  - 1}}<br />
{{1 + 49x^2 }}} dx \hfill \\

    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{1}<br />
{{98}}\left[ {\int {dx - } \int {\frac{1}<br />
{{1 + 49x^2 }}} dx} \right] \hfill \\

    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}<br />
{{98}} + \frac{1}<br />
{{98}}\int {\frac{1}<br />
{{1 + 49x^2 }}} dx \hfill \\

    = \frac{{x^2 }}<br />
{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}<br />
{{98}} + \frac{1}<br />
{{98 \times 7}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\

    = \frac{{x^2 }}{2}\tan ^{ - 1} \left( {7x} \right) - \frac{x}{{98}} + \frac{1}{{686}}\tan ^{ - 1} \left( {7x} \right) + c \hfill \\
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Integration - Dr. Evil is my teacher
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 9th 2009, 02:04 PM
  2. [SOLVED] Integration by Parts
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 1st 2009, 07:04 AM
  3. [SOLVED] Integration by Parts Help
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2009, 03:56 PM
  4. Integration by parts using arctan
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 28th 2009, 05:02 PM
  5. [SOLVED] Integration by parts
    Posted in the Calculus Forum
    Replies: 4
    Last Post: August 31st 2008, 03:15 PM

Search Tags


/mathhelpforum @mathhelpforum