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Integral check

  1. Jul 19, 2009 #1
    Hey i have this integral √1+y^2-(cothφ)ydy(with the square root on consisting of 1+y^2. I evaluated it but I just wanna you guys to check and see if you think its good. For the radical part I used a trig substitution y=tanφ
    The integral of tanφ is
    ln |sec x|
    And for the (cothφ)y part I figured it would just be (cothφ)y^2/2 because the function is in terms of dy and I figured I just integrate with respect to y. So this gives me the final answer
    ln |sec x|-(cothφ)y^2/2 for my final. Does this look good to you guys? My only concern about this is since my y=tanφ would I have to substitute it for the the y after the cothφ?
    Or I might be thinking of this completely wrong. Need advice!!
     
  2. jcsd
  3. Jul 19, 2009 #2

    rock.freak667

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    [tex]\int ( \sqrt{1+y^2} -y coth \phi) dy[/tex]

    if that is your integral, then your final answer is wrong I believe.

    For

    [tex]\int \sqrt{1+y^2} dy[/tex]

    if y=tan[itex]\theta[/itex], then what is dy= ?
    Then you will need to integrate more than tan[itex]\theta[/itex]
     
  4. Jul 19, 2009 #3
    so my dy would be sec^2φ.
    So if i plug that in il get
    ln |sec x|-ycothφsec^2φ. So would i have to plug in my y=tanφ for my y in front of the cothφ? If not would it just be ln |sec x|-y^2/2cothφsec^2φ?
     
  5. Jul 19, 2009 #4

    rock.freak667

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    Well firstly it is not wise to put the substituting variable as a symbol in the the integrand.

    So if you put y=tan[itex]\theta[/itex], use [itex]\theta[/itex] as the new variable and not [itex]\phi[/itex]


    In your original post, -(cothφ)y2/2 this part is correct. It is the first anti-derivative that is wrong.

    so back to it now. Yes dy= sec[itex]^2 \theta \ d\theta[/itex]

    [tex]\int \sqrt{1+y^2} dy \equiv \int (\sqrt{1+tan^2\theta} )sec^2\theta d\theta[/tex]

    Now do you know an identity for [itex]1+tan^2 \theta[/itex] ?
     
  6. Jul 20, 2009 #5
    yes it would be sec^2φ. Then I would square that and it would become secφ then multiply it be the sec^2φ I had for my dy. So im basically integrating sec^3φ?
     
  7. Jul 20, 2009 #6

    rock.freak667

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    yep.
     
  8. Jul 20, 2009 #7
    thanks!!
     
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