posted by Andre-Steve, please reply .
Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help.
Calculus - Steve, Tuesday, January 12, 2016 at 12:45am
1/2 ∫ 1/(sin(x/2)cos(x/2))dx
u = sin(x/2)
du = 1/2 cos(x/2) dx
or, dx = 2/cos(x/2) du
Then you have
1/4 ∫1/u 2/(cos(x/2))dx
= 1/4 ∫ 1/u du
= 1/4 ln(sin(x/2)) + C
Now, we all know that
∫ csc(x)dx = -ln(cscx + cotx)
so what gives here?
= -1/4 ln(1/sin(x/2))
= -1/2 ln(1/sqrt(1-cosx))
gotta run, but I think if you manipulate things a bit and adjust the C it will work out to be the same.
I'll check in later to make sure.
I still don't really understand it. I graphed -ln(|cscx + cotx|) and the answer you got, but they weren't the same. Thanks.
Well, I made a blunder, but you should have caught it.
dx = 2/cos(x/2) du
so, the integral becomes
∫ 1/2sin(x/2) * 1/cos(x/2) * 2/cos(x/2) du
= ∫ 1/(u(1-u^2)) du
= log(u) - (1/2)log(1-u^2)
I feel that is headed somewhere we want.
As usual, check my math.