# What is the arclength of the polar curve f(theta) = 2tan^2theta+cos3theta  over theta in [0,(5pi)/12] ?

Mar 9, 2016

${\int}_{0}^{\frac{5 \pi}{12}} \sqrt{{\left(3 \sin \left(3 t\right) - 4 {\sec}^{2} \left(t\right) \tan \left(t\right)\right)}^{2} + {\left(\cos \left(3 t\right) + 2 {\tan}^{2} \left(t\right)\right)}^{2}} \mathrm{dt} \approx 27.41370155 \ldots$

#### Explanation:

The Arc Length L in polar is given by:
L = int_a^b sqrt(r^2 + ((dr)/(d theta ))^2 d theta
where $a , b \in \left[0 , 2 \pi\right]$
For this case we need evaluate
$L = {\int}_{0}^{\frac{5 \pi}{12}} \sqrt{{f}^{2} \left(\theta\right) + {\left(\frac{\mathrm{df} \left(\theta\right)}{d \theta}\right)}^{2}} d \theta$
This a a nasty integral indeed:
${f}^{2} \left(\theta\right) = 4 {\tan}^{4} \theta + 4 {\tan}^{2} \theta \cos 3 \theta + {\cos}^{2} \theta$
$\frac{\mathrm{df} \left(\theta\right)}{d \theta} = 4 {\sec}^{2} \theta \tan \theta - 3 \sin 3 \theta$
${\left[\frac{\mathrm{df} \left(\theta\right)}{d \theta}\right]}^{2} = 16 {\left({\sec}^{2} \theta \tan \theta\right)}^{2} - 24 {\sec}^{2} \theta \tan \theta \sin 3 \theta + 9 \sin \left(3 \theta\right)$
Now add take the square root and compute using one of the online integral calculators.

${\int}_{0}^{\frac{5 \pi}{12}} \sqrt{{\left(3 \sin \left(3 t\right) - 4 {\sec}^{2} \left(t\right) \tan \left(t\right)\right)}^{2} + {\left(\cos \left(3 t\right) + 2 {\tan}^{2} \left(t\right)\right)}^{2}} \mathrm{dt} \approx 27.41370155 \ldots$

I use Wolframalpha