Arc length of $\sin^3({\frac x3})$ from $0$ to $6\pi$

Asked Oct 30 '13 at 09:28

Active Oct 30 '13 at 11:13

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How can I find the arc length for $P(x) = \sin^3\left({\frac x3}\right)$ from $0$ to $6\pi$.

I try to find:

$$\int_0^{6\pi} \sqrt{1 + \cos^2\left({\frac x3}\right)\sin^4\left({\frac x3}\right)}\,\mathrm dx$$

but I can't. Help me please, or just show me the way to solve it.

Maybe it could be done in polar coordinates?

edited Oct 30 '13 at 11:13

Lord_Farin

asked Oct 30 '13 at 09:28

Alexander

Maple cannot do that integral as an antiderivative. maybe with the specific bounds it can be done. – coffeemath Oct 30 '13 at 10:15
Have you tried using standard substitution? – AlexR Oct 30 '13 at 11:16

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