1. ## integral help

The problem is:

int (1+sinx)/(1-sin)dx

I tried doing this:
Let u=sinx, x=sin^(-1)u, so dx= du/sqrt(1-u^2)
Then I got
integral of (1+u)/(1-u)*(du)/(sqrt(1-u^2))
which is equal to
integral of ((1+u)*sqrt(1-u^2))/((1-u)^2(1+u)) du

I tried to figure it out using partial fractions from there, but that didn't work out. Any suggestions? Thanks!

When I multiply by (1-sinx)/(1-sinx) I just end up with
integral of cosx/(2-2sinx-cos^2x)dx
which I still don't know how to integrate.

2. Try multiplying either by (1-sinx)/(1-sinx).

3. When I multiply by (1-sinx)/(1-sinx) I just end up with
integral of cosx/(2-2sinx-cos^2x)dx
which I still don't know how to integrate.

4. Sorry meant (1+sinx)/(1+sinx). That will give you 1-sin^2(x)=cos^2(x) in the denominator and sin^2(x)+2sinx+1 in the numerator. Break it up into 3 fractions.

5. Just to verify-- did you solve it all the way through for an answer? I just want to make sure I did the rest correctly. I ended up with:

2tanx+2secx-x+C

Let me know if that looks right.
Thanks a lot!!

6. $\frac{ 1 + \sin(x)}{1 - \sin(x)} = \frac{ ( \sin(x/2) + \cos(x/2) )^2 }{ ( \sin(x/2) - \cos(x/2))^2}$

$= \tan^2(x/2 + \pi/4)$