Simple integral

  • Thread starter zoxee
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  • #1
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##\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
 

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  • #2
vela
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Just differentiate your answer and see if you recover the integrand.
 
  • #3
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Just differentiate your answer and see if you recover the integrand.
yes done that no problem,

just curious on how wolframalpha got their answer is all
 
  • #4
vela
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What did Wolfram Alpha give you?
 
  • #5
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##\dfrac{2sin(x/2)}{sin(x/2) + cos(x/2)}##
 
  • #6
vela
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Hint:
$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$
 
  • #7
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Hint:
$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$
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
 
  • #8
Ray Vickson
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##\dfrac{2sin(x/2)}{sin(x/2) + cos(x/2)}##
[tex] \frac{-2}{1+\tan(u)} = \frac{2\sin(u)}{\sin(u)+\cos(u)} + \text{constant}[/tex]
 

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