Calculate the lenght of $\vec{f}(t)=(\cos^4(t),\sin^4(t))$ from $0$ to $2\pi$

Question

$\vec{f}:[0,2\pi] \to \mathbb{R^2} $ ; $ \vec{f}(t)=(\cos^4(t),\sin^4(t))$.

Here is what I made so far...

$$ \begin{align} \ell( \overrightarrow{f} ) &= \int_0^{2\pi} \lVert \vec{f}\,' (t)\rVert \ dt \\ &= \int_0^{2\pi} \sqrt{16\cos^6(t)\sin^2(t)+16\cos^2(t)\sin^6(t)} \ dt \\ &=4 \int_0^{2\pi} |\cos(t)\sin(t)| \sqrt{\cos^4(t)+(1-\cos^2(t))^2} \ dt \\ &= 4 \int_0^{2\pi} |\cos(t)\sin(t)| \sqrt{2\left( \cos^2(t)-\frac{1}{2} \right)^2-1} \ dt\end{align}$$

From here I'm not sure what to do, I need some help. If you reply, please write your response as detailed as you can, if possible step by step, because I still do not handle so well this integrals. Thank you so much!

Note that $2\cos^2 t-1=\cos 2t$ and $\sin t\cos t=\frac{1}{2}\sin 2t$, so we will be making the preliminary substitution $u=\cos 2t$. The completing the square was imperfectly done, but the fix is not hard. — André Nicolas, Apr 08 '15 at 01:32

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

You found $$\| f' \|(t) = 4 \cdot \frac{|\sin(2t)|}{2} (\cos^4(t) + \sin^4(t))^{1/2}.$$ Since $$\cos^2(t) + \sin^2(t) = 1$$ we have that $$\cos^4(t) + \sin^4(t) = 1 - 2 \sin^2(t) \cos^2(t) = 1 - \frac{\sin^2(2t)}{2}.$$ This can be further simplified by noting that $$1 - \frac{\sin^2(2t)}{2} = \frac{1 + (1-\sin^2(2t))}{2} = \frac{1+\cos^2(2t)}{2}.$$ The integral takes the form

$$L = 2 \int_0^{2\pi} |\sin(2t)| \left( \frac{1+\cos^2(2t)}{2} \right)^{1/2} \, dt.$$

Perform the substitution $2t=u$ to find

$$L = \int_0^{\pi} |\sin(u)| \left( \frac{1+\cos^2 (u)}{2} \right)^{1/2} \, du.$$

On the interval $[0,\pi]$ we have $|\sin(u)| = \sin(u)$. Perform a second substitution $\cos(u) = \alpha$. Then $d \alpha = - \sin(u) \, du $ and $-1 \leq \alpha \leq 1$. The length will be

$$L = \int_{-1}^1 \frac{\sqrt{1+\alpha^2}}{\sqrt{2}} \, d \alpha.$$

Take a deep breath.

Perform a third substitution $\alpha = \tan(\theta)$. Then $d \alpha = \sec^2 (\theta) \, d \theta$ and $\sqrt{1+\alpha^2} = \sec(\theta)$, and so

$$L = \frac{1}{\sqrt{2}} \int_{- \pi/4}^{\pi/4} \sec^3 (\theta) \, d \theta.$$

Use that $$\int \sec^3(\theta) \, d \theta = \frac{1}{2} (\sec(\theta) \tan(\theta) + \ln|\sec(\theta) + \tan(\theta)|)$$ (see here) to obtain

$$\begin{align} L & = \frac{1}{2 \sqrt{2}} \left(2 \sqrt{2} +\ln \left( 3+2 \sqrt{2} \right) \right) \\ & = 1 + \frac{1}{2\sqrt{2}} \ln \left( 3+2\sqrt{2} \right). \end{align}$$

I don't see how $\cos^4(t) + \sin^4(t) = 1 - 2 \sin^2(t) \cos^2(t)$ — Карпатський, Apr 08 '15 at 15:35
@André Since $$\cos^2 (t) + \sin^2(t) = 1$$ then $$(\cos^2(t) + \sin^2(t))^2 = \cos^4 + 2 \sin^2(t) \cos^2(t) + \sin^4(t) = 1^2 = 1.$$ Thus $$\cos^4(t) + \sin^4(t) = 1 - 2 \sin^2(t) \cos^2(t).$$ — Mark Fantini, Apr 09 '15 at 02:41
Mark, thank you very much for your reply, it was very helpful. — Карпатський, Apr 09 '15 at 16:01

Calculate the lenght of $\vec{f}(t)=(\cos^4(t),\sin^4(t))$ from $0$ to $2\pi$

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