What's the integral of: 1/[1-cos(2x)]

Could you please explain the process. I did: [integral] 1 - sec(2x); is that correct?

No, it's wrong.

First, we have to reach an expression that's easier to integrate. We do this by using trigonometric identities.

cos(2x) = 1 - 2sin^2(x)      [trig identity]
1 - cos(2x) = 2sin^2(x)

int 1/[1-cos(2x)] dx = int 1/[2sin^2(x)] dx = (1/2) int 1/sin^2(x) dx

Now, that seems like an easier integral. Can you do it?

[quote author=Yoni link=topic=15642.msg157575#msg157575 date=1157567747]
No, it's wrong.

First, we have to reach an expression that's easier to integrate. We do this by using trigonometric identities.

cos(2x) = 1 - 2sin^2(x)      [trig identity]
1 - cos(2x) = 2sin^2(x)

int 1/[1-cos(2x)] dx = int 1/[2sin^2(x)] dx = (1/2) int 1/sin^2(x) dx

Now, that seems like an easier integral. Can you do it?
[/quote]

heh I converted 1/sin^2(x) to what is above without realizing that it's just csc^2(x) and you can find the integral to that easily. Thank you.