Arc Length for Parametric Equations $x={{\left( \cos u \right)}^{4}}$ and $y={{\left( \sin u \right)}^{4}}$

Question

We know $-1\le \cos u\le 1$, $0\le {{\left( \cos u \right)}^{4}}\le 1$, therefore, $0\le u\le \frac{\pi }{2}$. The length itself can be calculated as

$\begin{align} & \int\limits_{0}^{\frac{\pi }{2}}{\sqrt{{{\left( \frac{dx}{du} \right)}^{2}}+{{\left( \frac{dy}{du} \right)}^{2}}}du} \\ & =\int\limits_{0}^{\frac{\pi }{2}}{\sqrt{{{\left( -4\sin u{{\left( \cos u \right)}^{3}} \right)}^{2}}+{{\left( 4{{\left( \sin u \right)}^{3}}\cos u \right)}^{2}}}du} \\ & =4\int\limits_{0}^{\frac{\pi }{2}}{\sqrt{{{\left( \sin u \right)}^{2}}{{\left( \cos u \right)}^{6}}+{{\left( \sin u \right)}^{6}}{{\left( \cos u \right)}^{2}}}du} \\ \end{align}$

Unfortunately, I’m stuck with this integral. A hint would be very helpful. There was a similar question several years ago. Arc length of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$.

Answer 1

Simplify the integrand using the double angle formulae to get

$$I=\int{\sqrt{{{\left( \frac{dx}{du} \right)}^{2}}+{{\left( \frac{dy}{du} \right)}^{2}}}\,du} =\int \sqrt{\sin ^2(2 u) (\cos (4 u)+3)}\,du$$ In the range of integration, $\sin(2u)\geq 0 $; so $$I=\int\sin (2 u)\sqrt{3+\cos (4 u)}\,du$$ Using $\sin(2u)=t$, this gives $$I=\frac 1{\sqrt 2} \int t\, \sqrt{\frac{2-t^2}{1-t^2}}\,dt$$ Let $t^2=x$ to make $$I=\frac 1{2\sqrt 2} \int \sqrt{\frac{2-x}{1-x}}\,dx$$ which looks to be more pleasant.

Now, to finish, make $\sqrt{\frac{2-x}{1-x}}=w$

Answer 2

As a starter, and i'm not so sure how you may progress \begin{align} \sqrt{\sin^2 u \cos^6 u + \sin^6 u \cos^2 u} &= |\sin u||\cos u|\sqrt{\cos^4 u + \sin^4 u}\\ &=|\sin u||\cos u|\sqrt{(\sin^2 u + \cos^2 u)^2-2 \sin^2 u \cos^2 u}\\ &=|\sin u||\cos u|\sqrt{1-\frac{1}{2}(2 \sin u \cos u)^2}\\ &=|\sin u||\cos u|\sqrt{1-\frac{1}{2}\sin^2 2u} \end{align}

Answer 3

Rewrite the integral as

$$4\int_0^{\frac{\pi}{2}}\sin t\cos t\sqrt{\cos^4t+\sin^4t+2\cos^2t\sin^2t-2\cos^2t\sin^2t}\;dt$$

$$= \sqrt{2}\int_0^{\frac{\pi}{2}} \sin 2t \sqrt{1+\cos^22t}\:dt $$

This integral could be evaluated in many ways, but the simplest substitution would be $\cos 2t = \sinh \tau$ which gives

$$ \sqrt{2}\int_0^{\sinh^{-1}( 1)}\cosh^2\tau \:d\tau = \frac{1}{\sqrt{2}}\int_0^{\sinh^{-1}(1)} 1+ \cosh 2\tau \:d\tau = 1+\frac{\sinh^{-1}(1)}{\sqrt{2}}$$