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
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.
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.
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)
To integrate 1/sinx using the identity sinx=2(sin(x/2)cos(x/2)), you correctly rewrote the integral as 1/2 ∫ 1/(sin(x/2)cos(x/2))dx.
Now, let's continue:
Let u = sin(x/2)
Then, du = 1/2 cos(x/2) dx
Therefore, dx = 2/cos(x/2) du
Using this substitution, we can rewrite the integral as:
1/4 ∫ 1/u * 2/(cos(x/2)) du
= 1/4 ∫ 1/u du
Integrating 1/u with respect to u gives us ln|u| + C, where C is the constant of integration.
So, we have:
1/4 ln|u| + C
= 1/4 ln|sin(x/2)| + C
Now, let's compare this result to the standard integral of csc(x), which is -ln(|cscx + cotx|):
-1/4 ln|csc(x/2) + cot(x/2)| + C'
At first glance, it may seem like these two expressions are different. However, there is a relationship between the two.
Using the Pythagorean Identity: sin^2(x/2) + cos^2(x/2) = 1,
we have: sin(x/2) = √(1-cos(x/2)).
So, we can rewrite 1/4 ln|sin(x/2)| as:
-1/4 ln|1/sqrt(1-cos(x/2))|
With a bit of simplification, we get:
-1/2 ln|1/sqrt(1-cos(x))|
Therefore, when appropriately adjusted, the expression 1/4 ln|sin(x/2)| is indeed equivalent to the standard integral -ln(|cscx + cotx|).
I apologize for the confusion in my previous response. The two expressions should indeed be the same. If you graph both -ln(|cscx + cotx|) and -1/4 ln|sin(x/2)|, they should overlap. Please let me know if you have any further questions.