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.

Question

Calculus - Steve, Tuesday, January 12, 2016 at 12:45am
1/2 ∫ 1/(sin(x/2)cos(x/2))dx
let
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(sin(x/2))
= -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.

Answer 1

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)
= (1/2)log((u^2/(1-u^2))
I feel that is headed somewhere we want.

As usual, check my math.

Answer 2

1/2 log(u^2/(1-u^2))
1/2 log(sin^2(x/2)/cos^2(x/2))
log(tan(x/2))
-log(cot(x/2))
-log((1+cosx)/sinx)
-log(cscx+cotx)