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Integration problem 2

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx[/itex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]\int \frac{sin2x}{1 + cos^{2}x} \textrm{ } dx \\\\
    \int \frac{2sinxcosx}{1 + cos^{2}x} \textrm{ } dx \\\\
    u = cosx \\\\
    -\int \frac{2u}{1 + u^{2}} \textrm{ } du \\\\
    w = u^{2} \\\\
    -\int \frac{1}{1 + w} \textrm{ } dw \\\\
    q = 1 + w \\\\
    -\int \frac{1}{q} \textrm{ } dq \\\\
    -ln(q) = -ln(1 + w) = -ln(1 + u^{2}) = -ln(1 + cos^{2}x)[/itex]

    I don't know what I did wrong
     
  2. jcsd
  3. Feb 20, 2013 #2

    Curious3141

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    Your answer is correct, except for the omission of the constant of integration. Which makes all the difference in the world, really.

    Here's an easier way to approach the integral. Note that the denominator is, in fact: ##\frac{1}{2}(3 + \cos 2x)##. Can you go from there?

    You'll find that the final answer you get using that method is different from your answer by only a constant (##\ln 2##), which means the answers are equivalent.
     
  4. Feb 20, 2013 #3

    SammyS

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    What's the derivative of a constant?
     
  5. Feb 20, 2013 #4
    I don't see where that denominator is. The solution says it's: -ln(cos(2x) + 3) + C
     
  6. Feb 20, 2013 #5

    SammyS

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    What is [itex]\displaystyle \ \ -\ln(1 + \cos^{2}x)-(-\ln(\cos(2x) + 3))\ ?[/itex]
     
  7. Feb 20, 2013 #6
    Isn't it [itex]ln(\frac{1 + cos^2x}{cos(2x) + 3})[/itex]? But what does [itex]ln(cos(2x) + 3)[/itex] have to do with anything?
     
  8. Feb 20, 2013 #7

    Mute

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    Try plotting that function.
     
  9. Feb 20, 2013 #8
    It looks like -ln(2)
     
  10. Feb 20, 2013 #9

    SammyS

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    I get ln(2) .

    That's just a constant, so the two answers are equivalent.
     
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