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.
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.
To integrate the expression 1/sinx, you're on the right track in using the identity sinx = 2(sin(x/2)cos(x/2)).
You have rewritten the integral as 1/2 ∫ 1/(sin(x/2)cos(x/2))dx. Now let's simplify the expression further.
Recall the trigonometric identity: 1/(sinθcosθ) = 2/cos(2θ). In our case, since θ = x/2, we can rewrite the integral as:
1/2 ∫ 2/cos((x/2)x2) dx.
Now, observe that cos(2θ) = cos²(θ) - sin²(θ). Substituting x/2 for θ, we get:
cos(x) = cos²(x/2) - sin²(x/2).
So the integral becomes:
1/2 ∫ 2/(cos(x/2)² - sin(x/2)²) dx.
Using the trigonometric identity cos²(θ) - sin²(θ) = cos(2θ), we can further simplify the integral:
1/2 ∫ 2/cos(2(x/2)) dx.
Now, cos(2(x/2)) = cos(x), so the integral becomes:
1/2 ∫ 2/cos(x) dx.
At this point, we can integrate 2/cos(x) by using a trigonometric substitution.
Let's substitute cos(x) = t, which implies -sin(x) dx = dt.
The integral becomes:
1/2 ∫ -2/t dt.
Integrating -2/t gives you -2ln|t|. So the result is:
-1 ln|cos(x)| + C,
where C represents the constant of integration.
Therefore, integrating 1/sinx using the identity sinx = 2(sin(x/2)cos(x/2) yields -ln|cos(x)| + C.