Simple integral

1. Sep 10, 2013

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

2. Sep 10, 2013

vela

Staff Emeritus

3. Sep 10, 2013

zoxee

yes done that no problem,

just curious on how wolframalpha got their answer is all

4. Sep 10, 2013

vela

Staff Emeritus
What did Wolfram Alpha give you?

5. Sep 10, 2013

zoxee

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

6. Sep 10, 2013

vela

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

7. Sep 10, 2013

zoxee

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. Sep 10, 2013

Ray Vickson

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