Question

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

Answer 1

`int_0^((5 pi)/12) sqrt((3 sin(3 t)-4 sec^2(t) tan(t))^2+(cos(3 t)+2 tan^2(t))^2) dt~~27.41370155...`

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 [0, 2pi]`
For this case we need evaluate
`L = int_0^((5pi)/12) sqrt(f^2(theta ) + ((df(theta))/(d theta))^2 ) d theta`
This a a nasty integral indeed:
`f^2(theta)= 4tan^4theta + 4tan^2thetacos3theta +cos^2theta`
`(df(theta))/(d theta) = 4sec^2theta tantheta-3sin3theta`
`[(df(theta))/(d theta)]^2 = 16(sec^2theta tantheta)^2-24sec^2theta tanthetasin3theta + 9sin(3theta)`
Now add take the square root and compute using one of the online integral calculators.

`int_0^((5 pi)/12) sqrt((3 sin(3 t)-4 sec^2(t) tan(t))^2+(cos(3 t)+2 tan^2(t))^2) dt~~27.41370155...`

I use Wolframalpha