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

Question

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??

Answer 1

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 $21.25629$.

Answer 2

Hint

To compute $$I=\int \sec^3(x)\,dx$$ the tangent half-angle substitution seems to be convenient.

$$ t=\tan(\frac x2) \qquad dx=\frac{2\, dt}{1+t^2}\qquad \sec(x)=\frac{1+t^2}{1-t^2}$$ make $$I=2\int\frac{ \left(1+t^2\right)^2}{\left(1-t^2\right)^3}\,dt$$ Now, partial fraction decomposition leads to very simple terms.