I have to calculate the length bow $$r=4\sin^4\left(\frac{x}4\right), \space x∈(0,\pi)$$I know the formula that $$D= \int^\pi_0 { \sqrt{1+(r^*)^2}} dx$$The $r^*$ means the derivative of $r$ so $$r^*=4\sin^3\left(\frac{x}4\right)\cos\left(\frac{x}4\right)$$ So I have $$D=\int^\pi_0{ \sqrt{1+16\sin^6\left(\frac{x}4\right)\cos^2\left(\frac{x}4\right)}\space dx},$$but i don't know how to calculate this integral.
Have anyone any idea what I should do?
Thanks in advance!
$\endgroup$ 21 Answer
$\begingroup$First case
Given the arc of plane curve $\mathbf{r} : [a,\,b] \to \mathbb{R}^2$ of law:
$$ \mathbf{r}(t) := \left(t, \; f(t)\right), $$
the measure of its support $\gamma$ is equal to:
$$ ||\gamma|| = \int_{\gamma} |\mathbf{r}'(t)|\,\text{d}t = \int_a^b \sqrt{1 + {f'}^2(t)}\,\text{d}t\,. $$
In the specific case, we have $[a,\,b] = [0,\,\pi]$ and $f(t) = 4\,\sin^4\left(\frac{t}{4}\right)$, so we get:
$$ ||\gamma|| = \int_0^{\pi} \sqrt{1+16\sin^6\left(\frac{t}{4}\right)\cos^2\left(\frac{t}{4}\right)}\,\text{d}t \approx 3.42\,, $$
integral approximable only numerically.
Second case
Given the arc of plane curve $\mathbf{r} : [a,\,b] \to \mathbb{R}^2$ of law:
$$ \mathbf{r}(t) := \rho(t)\left(\cos t, \; \sin t\right), $$
the measure of its support $\gamma$ is equal to:
$$ ||\gamma|| = \int_{\gamma} |\mathbf{r}'(t)|\,\text{d}t = \int_a^b \sqrt{\rho^2(t) + {\rho'}^2(t)}\,\text{d}t\,. $$
In the specific case, we have $[a,\,b] = [0,\,\pi]$ and $\rho(t) = 4\,\sin^4\left(\frac{t}{4}\right)$, so we get:
$$ \begin{aligned} ||\gamma|| & = \int_0^{\pi} 4\,\sin^3\left(\frac{t}{4}\right)\text{d}t \\ & = \int_0^{\frac{\pi}{4}} 16\,\sin^3(u)\,\text{d}u \\ & = \int_0^{\frac{\pi}{4}} 16\left(\cos^2(u)-1\right)\left(-\sin(u)\,\text{d}u\right) \\ & = \int_1^{\frac{\sqrt{2}}{2}} 16\left(v^2-1\right)\text{d}v \\ & = \frac{4}{3}\left(\,8-5\sqrt{2}\,\right), \end{aligned} $$
where, in this case, the integral can be calculated analytically.
Note
Beyond the computational difficulties of the first case with respect to the second, I'm inclined to the second case since the exercise requires the length of a plane curve assigned in polar form $\rho = \rho(t)$, with $t \in [a,\,b]$.
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