Integrating a trig function divided by a trig function

[podcastube]

Homework Statement

Find the arc length of the curve r=4/θ, for ∏/2 ≤ θ ≤ ∏

Homework Equations

L= ∫ ds = ∫ √(r^2 + (dr/dθ)^2) dθ

The Attempt at a Solution

After some calculations, and letting θ = tanx, I now have to find ∫ ((secx)^3/(tanx)^2). I am not sure how to do this, but i have found online that ∫ (secx)^3 = (1/2)sec(x)tan(x)+(1/2)ln|sec(x)+tan(x)|. Using integration by parts and letting u = 1/(tan(x))^2= (cot(x))^2 and du/dx = -2cot(x).(cosec(x))^2 is looking very tedious. How do I solve this problem correctly?

 

[HallsofIvy]

I don't see how you got $sec^3(x)$. I get just sec(x) in the numerator.

Eventually, I reduce it to
$$\int \frac{sin^3(x)}{cos^2(x)}dx$$

Factor out a sin(x) to use with the dx and convert the remaining $sin^2(x)$ to $1- cos^2(x)$.