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Not sure about this integral

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data
    [tex]\int \frac{dx}{(x^{2}+2x+2)^{2}}[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I'm just gonna skip down to where something is wrong (the beginning is a u-substitution and trig. substitution so I don't want to type all that out)
    After all the trig/u-subs I end up with:
    [tex]\int cos^{2}\theta d \theta=\int \frac{1}{2}+\frac{cos2 \theta}{2}d \theta=\frac{\theta}{2}+\frac{sin2\theta}{4}+C[/tex]

    So now I substitute back, u=tanθ:
    [tex]\frac{tan^{-1}u}{2}+\frac{u}{2\sqrt{u^{2}+1}}[/tex]
    and u=x+1:
    I THINK THE ERROR IS HERE SOMEWHERE, BUT NOT SURE WHERE
    [tex]\frac{1}{2}(tan^{-1}(x+1)+\frac{x+1}{\sqrt{(x+1)^{2}+1}})+C[/tex]


    Thank you
     
  2. jcsd
  3. Feb 13, 2013 #2
    Your error is when you substitute ##\theta## back into ##\displaystyle \frac{sin2\theta}{4} \not = \frac{u}{2\sqrt{u^{2}+1}}##
     
    Last edited: Feb 13, 2013
  4. Feb 13, 2013 #3
    If you draw the triangle where u=tanθ, then sine of that would be [itex]\frac{u}{\sqrt{u^{2}+1}}[/itex] wouldn't it? And then multiplying that by 2 would just be 2 times that, but I think that's the error, I'm not sure what the 2 does to the sine function.
     
  5. Feb 13, 2013 #4
    No, ##sin(2arctan(x)) \not = 2sin(arctan(x))##. I don't know how to derive ##sin(2arctan(x))## but a Google search could help.

    Instead of finding an expression without trig functions, sometimes it's just best to leave it as is. The answer is still perfectly valid.
     
  6. Feb 13, 2013 #5
    Ok yeah I see that's the problem, but even after googling it, that doesn't help :\
     
  7. Feb 14, 2013 #6
    Still don't understand this, could someone explain how to get sin(2arctan(x))?
     
  8. Feb 14, 2013 #7

    Mute

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    Homework Helper

    Use ##\sin(2\theta) = 2\sin \theta \cos \theta##, then write the sin and cos in terms of u.
     
  9. Feb 14, 2013 #8
    Wow forgot that, thanks so much.
     
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