Find the area of the surface for the curve rotated about the x-axis 
$y = \tan(x), 0\leq x \leq \frac{\pi}{3}$

I am struggling with constructing an integral for this formula.
Since the curve is rotated about the x-axis, I think this is the best way to setup the integral. Since the circumference is $2\pi r$, it makes sense to me that the radius is $y=f(x)$. 
For the length of the curve,
$$ds = \sqrt{1 + \bigg(\frac{dy}{dx}\bigg)^2}dx = \sqrt{1 + \bigg(\frac{dx}{dy}\bigg)^2}dy $$
Since I can find $\frac{dy}{dx}$ relatively easily, $\frac{dy}{dx} = \sec^2x$, it would seem easiest for me to set up the integral as follows:
$$ S = \int 2\pi \cdot y \cdot ds = \int 2\pi \cdot y \cdot \sqrt{1+\frac{dy}{dx}^2}dx$$
Please let me know if my thinking is correct up to this point.
$$\begin{align}
S &= \int 2\pi y ds \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+(\sec^2x)^2}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+\sec^4x}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \tan x \sqrt{1+ \frac{1}{\cos^4x}}dx \\
&= 2\pi\int_{0}^{\frac{\pi}{3}} \frac{\sin x}{\cos x} \frac{\sqrt{1+cos^4x}}{\cos^2x}dx \\
&= 2\pi\int_{\frac{1}{2}}^{1} \frac{\sqrt{1+u^4}}{u^3}du
\end{align}$$
Substitution - $u = \cos x, -du = \sin x $
Edit: Here is where I am getting stuck 
Edit: Edit: I am giving up on this problem. I just did a parts by substitution and then looks like I have to do another substitution + trigonometric substitution to solve that integral......My variables are a complete mess at this point.

Hint: Do a change of variable $\cos x=u$, then $du=-\sin x$, so $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+x^2}}{x^2}du$$
Note: Since the derivative will need to be squared,  $$S=2\pi\int_{1/2}^1\frac{\sqrt{1+u^4}}{u^3}du$$
Here are some hints how to solve the rest: Make substitution $v=u^2$, $dv=2udu$ and you get an integral$$\frac 12\int\frac{\sqrt{1+v^2}}{v^2}dv$$
Integrate by parts $f=\sqrt{1+v^2}$, $g'=\frac{1}{v^2}$, with $f'=\frac{v}{\sqrt{1+v^2}}$ and $g=-\frac 1v$. Then all you are left with is to calculate $$\int\frac 1{\sqrt{1+v^2}}dv$$
