this problem, I don't even have a clue where to start...

Substitute

 

sinx = 2t/(1+t^2)

 

cosx = (1-t^2)/(1+t^2)

 

where t = tan(x/2), dt/dx = 1/2*(1+t^2)

 

rearranging gives

 

/

| dt/(1+t) = ln(1 + tan(x/2)) + c

/

This one is kind of a mess.

Try integration by parts. Let [math]dv=dx[/math] [math]u=\frac{1}{1+cosx+sinx}[/math]

Do that process, then you'll have to do integration by parts again. Have fun.

Edit: Nevermind! This just brings you in circles. Tartaglia's is probably better.

Substitute

 

sinx = 2t/(1+t^2)

 

cosx = (1-t^2)/(1+t^2)

 

where t = tan(x/2)' date=' dt/dx = 1/2*(1+t^2)

 

rearranging gives

 

/

| dt/(1+t) = ln(1 + tan(x/2)) + c

/[/quote']

I got the answer

/

| ln(2 + 2tan(x/2)) + c

/

 

after looking up my identity sheet and I saw those substitutions but for dt/dx it is (2dt) / (1+t^2) which is different from your 1/2*(1+t^2).

Your expression is for dx not dt/dx and so it is tha same as mine only upside down.

 

You are also obviously unfamiliar with the laws of logs as

 

ln(2 + 2tan(x/2)) = ln2 + ln(1+tan(x/2))

 

and the ln2 disappears into the integration constant

Your expression is for dx not dt/dx and so it is tha same as mine only upside down.

 

You are also obviously unfamiliar with the laws of logs as

 

ln(2 + 2tan(x/2)) = ln2 + ln(1+tan(x/2))

 

and the ln2 disappears into the integration constant

ahh, you're right, I am not familiar with logs, heh. and o yeah that thing on my sheet is dx. lol. thanks!!! you're a genius at this, you got my vote