# Simple integral

##\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
Staff Emeritus
Homework Helper

yes done that no problem,

just curious on how wolframalpha got their answer is all

vela
Staff Emeritus
Homework Helper
What did Wolfram Alpha give you?

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

vela
Staff Emeritus
Homework Helper
Hint:
$$\frac{\tan x}{1+\tan x} = \frac{\tan x + 1 - 1}{1+\tan x} = 1 - \frac{1}{1+\tan x}$$

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