Evaluating ∫[cotx/(1+sinx)]dx from 0 to π/2

[dying]

I’ve been stuck on this question for quite a while:

Evaluate ∫[cotx/(1+sinx)]dx from 0 to π/2

I think I’m supposed to substitute cotx to cosx/sinx, and I eventually ended up with ∫(cscxsecx - secx)dx, then ln|secx| + ln|sinx| - ln|secx+tanx|. I plopped the values in and got the wrong answer, the book says the answer is ln2. I’m very confused on what I did wrong and how to do it correctly, and would appreciate any help, thanks in advance!

 

[stapel]

I’ve been stuck on this question for quite a while:

Evaluate ∫[cotx/(1+sinx)]dx from 0 to π/2

I think I’m supposed to substitute cotx to cosx/sinx, and I eventually ended up with ∫(cscxsecx - secx)dx, then ln|secx| + ln|sinx| - ln|secx+tanx|. I plopped the values in and got the wrong answer, the book says the answer is ln2. I’m very confused on what I did wrong and how to do it correctly, and would appreciate any help, thanks in advance!

Please reply *showing* your steps and answer. ("Read Before Posting") Once we have that information, we can try to figure out where things are going sideways.

Please be complete. Thank you!

 

[skeeter]

[imath]\displaystyle \int_0^{\pi/2} \dfrac{\cos{x}}{\sin{x}(1+\sin{x})} \, dx[/imath]

use the substitution [imath]u = 1+\sin{x}[/imath], get an integrand in terms of [imath]u[/imath], and reset the limits of integration.

From there, I would recommend using a partial fraction decomposition.

 

[BobP]

I’ve been stuck on this question for quite a while:

Evaluate ∫[cotx/(1+sinx)]dx from 0 to π/2

I think I’m supposed to substitute cotx to cosx/sinx, and I eventually ended up with ∫(cscxsecx - secx)dx, then ln|secx| + ln|sinx| - ln|secx+tanx|. I plopped the values in and got the wrong answer, the book says the answer is ln2. I’m very confused on what I did wrong and how to do it correctly, and would appreciate any help, thanks in advance!

The integrand is infinite at the lower limit.
I think that there's a typo and it should be cosx on the top line, not cotx .

 

[pka]

The integrand is infinite at the lower limit.
I think that there's a typo and it should be cosx on the top line, not cotx .

That makes much more sense. It changes the integral to one that converges and is a simple logarithm.