Q:

Arc length of Archimedes Spiral $ r = \theta $ from $ 0 \le \theta \le 2\pi$

The equation of the Archimedes spiral is given by $$r = \theta$$
The formula for calculating the Arc Length is given by $$L = \int^b_a\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$$
The following is my work on this problem so far:
$$L = \int^{2\pi}_0\sqrt{\theta^2+1}d\theta$$
Using Trig substitution
$$\theta = \tan(u)$$
$$ d\theta = \sec^2(u)du$$
$$L = \int^{\tan^-1(2\pi)}_{\tan^-1(0)}\left(\sqrt{\tan^2(u)+1}\right)\sec^2(u)du$$
$$L = \int^{\tan^-1(2\pi)}_{\tan^-1(0)}\left(\sqrt{\sec^2(u)}\right)\sec^2(u)du$$
$$L = \int^{\tan^-1(2\pi)}_{\tan^-1(0)}\sec^3(u)du$$
This is basically where I am stuck, and I am also not sure if I am over complicating this problem or am I on the right track?
Ok so after using the integration by parts for the above integral I arrive at:
$$\left[\frac{\sec(u)\tan(u) + ln |\sec(u)+\tan(u)|}{2}\right]^{\tan^{-1}({2\pi})}_{\tan^{-1}(0)}$$
How to evaluate that without a calculator??

A:

You are almost done! Use 
$$\tan(\arctan(x))=x$$
(per definition) and
$$\sec(\arctan(x))=\sqrt{x^2+1}$$
(convince yourself of this by looking at a right-angled triangle for a few seconds) to arrive at the result
$$\pi\sqrt{4\pi^2+1}+\frac12\ln\left(\sqrt{4\pi^2+1}+2\pi\right)$$
There is probably no easy way to approximate that number without a calculator, but if I plugged the numbers correctly into mine it should be about $21.25629$.

