# Calculate the length of a curve!

I must calculate the lenght of this curve $$\phi(t)=\big(\sin(2t),\cos(3t)\big)$$ so I must to calculate the integral of the norm of the derivative $$\left \| D\phi(t) \right \|=\sqrt{4\cos^2(2t)+9\sin^2(3t)}$$

$$\int _0^{2\pi} \sqrt{4\cos^2(2t)+9\sin^2(3t)}\, dt$$

I tried with various method, to change the function until they have the same arguments, but I can't solve this! Do you have some hints?

• Mathematica can't return an exact answer... Maybe you meant the $\sin(3t)$ to be a $\sin(2t)$? Or were happy just to set up the integral? Or can settle for a numerical approximation (which is 15.2894 here)? Dec 1, 2014 at 17:46
• @JohnD yep I got 15,2... by wolfram alpha, but this is one exercise of my math exam, so I should know how to get this 15,28.. Dec 1, 2014 at 20:26
• If Mathematica can't integrate that, you won't either. Dec 1, 2014 at 21:07

$$\int _0^{2\pi} \sqrt{4\cos^2(2t)+9\sin^2(3t)}\, dt = \int _0^{2\pi} \sqrt{\frac{13}{2}}\sqrt{1 +\frac{4}{13}cos(4t)-\frac{2}{13}cos(6t)} dt=$$ $$=\sqrt{\frac{13}{2}}\int _0^{2\pi} 1 +\frac{1}{2}(\frac{4}{13}cos(4t)-\frac{2}{13}cos(6t))-\frac{1}{8}(\frac{4}{13}cos(4t)-\frac{2}{13}cos(6t))^2+.. dt\approx 15$$