integral from 1/18 to 1/6 of csc(3pit)cot(3pit)dt
To solve this integral, we can use the trigonometric identity:
csc(x) = 1/sin(x)
cot(x) = 1/tan(x)
Substituting these identity in the integral expression, we have:
∫[1/18 to 1/6] 1/(sin(3πt) * tan(3πt)) dt
Now, we can use a substitution to simplify the integral. Let u = 3πt, du = 3π dt.
When t = 1/18, u = 3π/18 = π/6.
When t = 1/6, u = 3π/6 = π/2.
Using the limits of integration in terms of u, the integral becomes:
∫[π/6 to π/2] (1/(sin(u)*tan(u))) * (du/(3π))
Simplifying the expression further:
1/(3π) * ∫[π/6 to π/2] (1/(sin(u)*tan(u))) du
Using the identity tan(x) = sin(x)/cos(x), the integral becomes:
1/(3π) * ∫[π/6 to π/2] (1/(sin(u)*sin(u)/cos(u))) du
Simplifying the expression further:
1/(3π) * ∫[π/6 to π/2] (cos(u) / (sin(u) * sin(u))) du
Using another trigonometric identity 1/(sin^2(x)) = csc^2(x), the integral becomes:
1/(3π) * ∫[π/6 to π/2] (cos(u) * csc^2(u)) du
Integrating the expression with respect to u:
1/(3π) * (-csc(u)) | from π/6 to π/2
Now, we can substitute the limits of integration:
1/(3π) * (-csc(π/2) + csc(π/6))
Since csc(x) = 1/sin(x), the expression simplifies to:
1/(3π) * (-1 + 2)
This further simplifies to:
-1/(3π) + 2/(3π)
Therefore, the final answer is:
-1/(3π) + 2/(3π) = (2 - 1)/(3π) = 1/(3π)
1/18 to 1/6 of csc(3pit)cot(3pit)dt
1/18 to 1/6 of [1/sin(3 pi t) ] cos(3pit) / [sin (3 pi t)] dt
1/18 to 1/6 of cos(3pit) / [sin^2 (3 pi t)] dt
if y = sin (3 pi t) then dy = 3 pi cos(3 pi t) dt
so
1/18 to 1/6 in t and [1/y^2] dy/ (3 pi)
if t= 1/18 then y = sin pi/6 = 1/2
if t = 1/6 then y = sin pi/2 = 1
so
you really have the integral from y = 1/2 to y = 1 of 1/(3 pi) times y^-2 dy
which is
1/2 to y = 1 of 1/(3 pi) times -1/y
if y = 1, -1/(3 pi)
if y = 1/2 , +2/(3 pi)
so 3/(3 pi) =1/(3pi)
so agree