The integral is not easy, but doable. I will not worry about constants. Make the change of variable $\phi=\theta/3$. Or not. Bring a $\sin\phi$ outside the square root. (Technically it should be $|\sin\phi|$, but if $\phi$ does not stray beyond $[0,\pi]$ we don't need to worry.) We are integrating $\sin\phi\sqrt{\sin^2\phi+(4/9)\cos^2\phi}.$ Inside the square root, replace $\sin^2\phi$ by $1-\cos^2\phi$. So we need to find the integral of $\sin\phi\sqrt{9-5\cos^2\phi}.$ Make the substitution $u=\cos\phi$. We end up needing to integrate $\sqrt{9-5u^2}.$ Now it is somewhat downhill. Let $u=\dfrac{3}{\sqrt{5}}v$.
With particular limits, things might be easier. For with appropriate limits the definite integral of $\sqrt{9-5u^2}$ will be the perhaps easily found area of a nice part of a circle.