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Math Help - integration of 3 variables

  1. #1
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    integration of 3 variables

    integral of <||cos t, sin t, t||> dt

    the answer in the back has a natural log and i don't get it when i work it out???
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by chris25 View Post
    integral of <||cos t, sin t, t||> dt

    the answer in the back has a natural log and i don't get it when i work it out???
    Is it ||...|| for the norm in basis 2 ?


    It would be ||\cos t, \sin t, t||=\sqrt{\cos^2t+\sin^2t+t^2}=\sqrt{1+t^2}
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  3. #3
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    no it means the length
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    Moo
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    Quote Originally Posted by chris25 View Post
    no it means the length
    Well, it has to be the same...

    Can you state and write the exact question ?
    Are there really < and > ?
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  5. #5
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    evaluate the indefinite integral...

    S||(cos t i + sin t j + t k )|| dt
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  6. #6
    Moo
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    Quote Originally Posted by chris25 View Post
    evaluate the indefinite integral...

    S||(cos t i + sin t j + t k )|| dt
    So yeah, you have to calculate \int \sqrt{1+t^2} dt

    But I'm not sure how to calculate it...
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    So yeah, you have to calculate \int \sqrt{1+t^2} dt

    But I'm not sure how to calculate it...
    \int\sqrt{1+x^2}dx

    let x=\tan(\theta)

    so dx=\sec^2(\theta)d\theta

    Giving us after simplification

    \int\sec^3(\theta)d\theta

    Which by a manipulation and double parts gives http://www.mathhelpforum.com/math-he...-integral.html

    \frac{1}{2}\bigg[\ln|\sec(\theta)+\tan(\theta)|+\tan(\theta)\sec(\t  heta)\bigg]

    Now back subbing in that \theta=\arctan(x)

    we get

    \frac{1}{2}\bigg[\ln|\sqrt{x^2+1}+x|+x\sqrt{x^2+1}\bigg]
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  8. #8
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    Quote Originally Posted by Mathstud28 View Post
    \int\sqrt{1+x^2}dx

    let x=\tan(\theta)

    so dx=\sec^2(\theta)d\theta

    Giving us after simplification

    \int\sec^3(\theta)d\theta

    Which by a manipulation and double parts gives http://www.mathhelpforum.com/math-he...-integral.html

    \frac{1}{2}\bigg[\ln|\sec(\theta)+\tan(\theta)|+\tan(\theta)\sec(\t  heta)\bigg]

    Now back subbing in that \theta=\arctan(x)

    we get

    \frac{1}{2}\bigg[\ln|\sqrt{x^2+1}+x|+x\sqrt{x^2+1}\bigg]
    The substitution x = \sinh \theta will reduce the algebra.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    The substitution x = \sinh \theta will reduce the algebra.
    Yeah, but I wanted to supply a more commonly known sub

    But to explain what Mr. F astutely noted

    Let x=\sinh(\theta)

    So dx=\cosh(\theta)

    Giving us

    \int\sqrt{1+\sinh^2(\theta)}\cosh(\theta)d\theta=\  int\cosh^2(\theta)d\theta

    Now making the appropriate hyperbolic identity we get

    \frac{1}{2}\bigg[\theta+\cosh(\theta)\sinh(\theta)\bigg]

    Now just back sub
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  10. #10
    Moo
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    Quote Originally Posted by mr fantastic View Post
    The substitution x = \sinh \theta will reduce the algebra.
    Actually, the OP mentioned that the answer contained a logarithm
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