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Math Help - arctan integral

  1. #1
    Junior Member Coco87's Avatar
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    arctan integral

    Hey,
    I am on a problem that involves integrating a arctan:

    \int{\arctan{\left(\frac{1}{x}\right)}}dx

    I think what is throwing me off is the \frac{1}{x}

    I set:

    u=\arctan{\left(\frac{1}{x}\right)}
    dv=dx

    so:

    du=\frac{1}{1+\left(\frac{1}{x}\right)^2}
    v=x

    and I get:

    x\arctan{\left(\frac{1}{x}\right)}-\int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx

    I think \int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx is the problem for me. No matter how I work it, I can't seem to get the answer (maple):

    x\arctan{\left(\frac{1}{x}\right)}+\frac{1}{2}\ln{  \left(1+\frac{1}{x^2}\right)}-\ln{\left(\frac{1}{x}\right)}

    Any ideas?
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  2. #2
    Super Member 11rdc11's Avatar
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    Try subbing

    \frac{1}{x} = \tan\theta
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  3. #3
    Junior Member Coco87's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Try subbing

    \frac{1}{x} = \tan\theta
    Thanks for the reply

    I'm not sure what you mean by subbing though

    I'm in Calc 2 on Integrating by Parts; not sure if I have been tought that technique yet, or maybe I've totally missed that part from past courses

    By subbing, do you mean replace all of the \frac{1}{x}'s with \tan{x}
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coco87 View Post
    Thanks for the reply

    I'm not sure what you mean by subbing though

    I'm in Calc 2 on Integrating by Parts; not sure if I have been tought that technique yet, or maybe I've totally missed that part from past courses

    By subbing, do you mean replace all of the \frac{1}{x}'s with \tan{x}
    he is talking about a trig substitution. chances are you learned that technique before learning the integration by parts technique

    anyway, you took the derivative of arctan(1/x) incorrectly. you need to use the chain rule. chances are that's what's messing you up.
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  5. #5
    Super Member 11rdc11's Avatar
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    Yep put \tan\theta = \frac{1}{x}

    Remember to change the other values to theta too

    x = \frac{1}{\tan\theta}

    dx = -\csc^2\theta~d\theta

    and use the identity

    1 + \tan^2\theta = \sec^2\theta

    so you get

    \int \frac{-\csc^2\theta}{\tan\theta\sec^2\theta}~d\theta
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  6. #6
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    Quote Originally Posted by Coco87 View Post
    Hey,
    I am on a problem that involves integrating a arctan:

    \int{\arctan{\left(\frac{1}{x}\right)}}dx

    I think what is throwing me off is the \frac{1}{x}

    I set:

    u=\arctan{\left(\frac{1}{x}\right)}
    dv=dx

    so:

    du=\frac{1}{1+\left(\frac{1}{x}\right)^2}
    v=x

    and I get:

    x\arctan{\left(\frac{1}{x}\right)}-\int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx

    I think \int{\frac{x}{1+\left(\frac{1}{x}\right)^2}}dx is the problem for me. No matter how I work it, I can't seem to get the answer (maple):

    x\arctan{\left(\frac{1}{x}\right)}+\frac{1}{2}\ln{  \left(1+\frac{1}{x^2}\right)}-\ln{\left(\frac{1}{x}\right)}

    Any ideas?
    To get du you have to use the chain rule: du = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \cdot {\color{red}\left( - \frac{1}{x^2}\right) } dx = \frac{-1}{x^2 + 1} \, dx.

    Integration by parts leads to a simple calculation.


    Note: All the ln stuff Maple gives you is equivalent to \frac{1}{2} \ln (x^2 + 1) \, ....
    Last edited by mr fantastic; September 27th 2008 at 03:12 PM. Reason: Added the note
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Jhevon View Post
    he is talking about a trig substitution. chances are you learned that technique before learning the integration by parts technique

    anyway, you took the derivative of arctan(1/x) incorrectly. you need to use the chain rule. chances are that's what's messing you up.
    Thanks i forgot to check his work.
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  8. #8
    Junior Member Coco87's Avatar
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    Sorry about the late response here, but I've been messing with the problem

    Thank you all for your help!

    I actually looked up trig substitution, and it's 2 sections after this one (hopefully it makes dealing with this stuff easier ). I also need to get use to maple, I got x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{  2}\right)\ln{\left|x^2+1\right|}, and I kept on comparing it to the answer maple got, and it kept showing me false. I finally just did the derivative of my answer, and then simplified it and got \arctan{\left(\frac{1}{x}\right)}. So then I played with the answer maple gave me and simplified it symbolically, and it gave me the same answer as I got.

    I guess I should RTFM

    lol, thanks again guys!
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Coco87 View Post
    Sorry about the late response here, but I've been messing with the problem

    Thank you all for your help!

    I actually looked up trig substitution, and it's 2 sections after this one (hopefully it makes dealing with this stuff easier ). I also need to get use to maple, I got x\arctan{\left(\frac{1}{x}\right)}+\left(\frac{1}{  2}\right)\ln{\left|x^2+1\right|}, and I kept on comparing it to the answer maple got, and it kept showing me false. I finally just did the derivative of my answer, and then simplified it and got \arctan{\left(\frac{1}{x}\right)}. So then I played with the answer maple gave me and simplified it symbolically, and it gave me the same answer as I got.

    I guess I should RTFM

    lol, thanks again guys!
    RTFM?

    anyway, Mr F told you that answer was equivalent
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