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Simple integral

  1. Sep 10, 2013 #1
    ##\displaystyle \int \dfrac{1}{1+sin(\theta)}\ d\theta ##

    I got ## -2(1+tan(\frac{\theta}{2}))^{-1} ## however wolframalpha got a different result than me, could anyone double check this for me as I've been over it a lot of times and can't get the wolframalpha equivalent
     
  2. jcsd
  3. Sep 10, 2013 #2

    vela

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    Just differentiate your answer and see if you recover the integrand.
     
  4. Sep 10, 2013 #3
    yes done that no problem,

    just curious on how wolframalpha got their answer is all
     
  5. Sep 10, 2013 #4

    vela

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    What did Wolfram Alpha give you?
     
  6. Sep 10, 2013 #5
    ##\dfrac{2sin(x/2)}{sin(x/2) + cos(x/2)}##
     
  7. Sep 10, 2013 #6

    vela

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    Hint:
    $$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$
     
  8. Sep 10, 2013 #7
    Can't seem to use that hint... what do I need to do to get from my step to their step? They are not equivalent as the constant of integration is different
     
  9. Sep 10, 2013 #8

    Ray Vickson

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    [tex] \frac{-2}{1+\tan(u)} = \frac{2\sin(u)}{\sin(u)+\cos(u)} + \text{constant}[/tex]
     
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