Curve length and very hard integral

[Mathitalian]

Homework Statement

Find the length of the curve:

$\phi(t)=\left\{(5+\cos(3t))\cos(t), (5+\cos(3t))\sin(t) \right\}\mbox{ with } t\in [0, 2\pi]$

Homework Equations

$L_{\phi}= \int_{a}^{b}\sqrt{[x'(t)]^2+ [y'(t)]^2}\qquad (1.1)$

Where

$x(t)= (5+\cos(3t))\cos(t)$

$y(t)= (5+\cos(3t))\sin(t)$

$a= 0\qquad b= 2\pi$

The Attempt at a Solution

Ok, i noticed that $\phi(t)$ is in this form:

$\phi(t)=(r(t)\cos(t), r(t)\sin(t))$


so it can be expressed in polar form:

$r= r(t)\iff r=5+\cos(3t) \quad t\in[0,2\pi]$

so:

$L_{\phi}=\int_{0}^{2\pi}\sqrt{[r(t)]^2+[r'(t)]^2}=$

$= \int_{0}^{2\pi}\sqrt{(5+\cos(3t))^2+(-3\sin(3t))^2}dt$


but this integral is not so easy to solve :(

What can i do to solve it? I try to use wolfram but it gives me an approximate result.

[Sorry, my English is not so good, forgive me if there are mistakes]

 

[CompuChip]

So far I don't see any errors.
Are you sure you need the result in closed form, and you're not allowed to do numeric integration?

Also, are you sure it's not
$$\phi(t) = \{ 5 + (3 \cos(3t) \cos(t)), 5 + (3 \cos(3t) \sin(t)) \}$$

 

[Mathitalian]

Thanks for your reply CompuChip :)

I can't use numeric integration, but if you tell that this integral can't be expressed in closed form I'll go to my teacher to check it.

 

[CompuChip]

I'm not sure there is no clever substitution that will do the trick. But there is no immediately obvious one; and if the best Wolfram Alpha does is also give you a numeric answer, that's a pretty strong hint that there is no "nice" form.

 

[Mathitalian]

Yes, you are right, i have the same feeling. Thank you for your time!