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 from: Yoni on September 06, 2006, 01:35 PM
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?
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.