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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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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

vela

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Just differentiate your answer and see if you recover the integrand.

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yes done that no problem,Just differentiate your answer and see if you recover the integrand.

just curious on how wolframalpha got their answer is all

- #4

vela

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What did Wolfram Alpha give you?

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##\dfrac{2sin(x/2)}{sin(x/2) + cos(x/2)}##

- #6

vela

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Hint:

$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$

$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$

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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 differentHint:

$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$

- #8

Ray Vickson

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

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