Simple integral

[zoxee]

$\displaystyle \int \dfrac{1}{1+sin(\theta)}\ d\theta$

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

 

[vela]

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

 

[zoxee]

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

 

[vela]

What did Wolfram Alpha give you?

 

[zoxee]

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

 

[vela]

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

 

[zoxee]

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

 

[Ray Vickson]

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

$$\frac{-2}{1+\tan(u)} = \frac{2\sin(u)}{\sin(u)+\cos(u)} + \text{constant}$$