Arc length of $f(t)=(cos^4(t),sin^4(t))$ from $t=0$, to $t=2\pi$. I'm trying to compute the arc length $L$ of $f(t)=(cos^4(t),sin^4(t))$  from $t=0$, to $t=2\pi$. Using the regular formula we have that:
$$L=\int_{0}^{2\pi}\sqrt{16\cos^2(t)\sin^2(t)(\cos^4(t)+\sin^4(t))}dt$$
Then after some (long) calculi it reduced to 
$$L=4\int_0^{\pi/2}\sin(2t)\sqrt{3+cos(4t)}dt$$
and then this  can be computed using integration by parts, which leads to the integral 
$$I=\int\frac{\sin(4t)\cos(2t)}{\sqrt{3+\cos(4t)}}dt$$
Now, by making $z=cos(2t)$ we have that 
$$I=\int\frac{z^2}{\sqrt{2z^2+2}}dz=-\frac{1}{\sqrt2}\int\tan^2\theta\sec\theta d\theta$$
taking $z=\tan\theta$. Finally, i can make the tangent a secant squared, which will make me compute the integral of a secant cube, which involves more integration by parts. The point is that this is a very long and tedious approach to compute this arc length. Can you suggest a simpler way of computing this number? Any idea will be helpful.

It is better to write $$\sin 2t \sqrt{3 + \cos 4t} = \sin 2t \sqrt{2 + 2 \cos^2 2t},$$ hence with the substitution $$u = \cos 2t, \quad du = -2 \sin 2t \, dt,$$ we obtain $$\int_{t=0}^{\pi/2} \!\!\sin 2t \sqrt{2 + 2 \cos^2 2t} \, dt = -\frac{1}{2} \int_{u=1}^{-1} \!\!\!\!\sqrt{2 + 2u^2} \, du = \frac{1}{\sqrt{2}} \int_{u=-1}^1 \!\!\!\! \sqrt{1 + u^2} \, du \\ = \sqrt{2} \int_{u=0}^1 \!\!\sqrt{1+u^2} \, du.$$  Then integration by parts yields 
$$\begin{align*}I = \int_{u=0}^1 \sqrt{1+u^2} \, du 
&= \left[ u \sqrt{1+u^2} \right]_{u=0}^1 - \int_{u=0}^1 \frac{u^2}{\sqrt{1+u^2}} \, du \\
&= \sqrt{2} - \int_{u=0}^1 \frac{1+u^2}{\sqrt{1+u^2}} - \frac{1}{\sqrt{1+u^2}} \, du \\
&= \sqrt{2} - I + \left[ \sinh^{-1} u \right]_{u=0}^1,
\end{align*}$$
consequently $$I = \frac{\sqrt{2} + \sinh^{-1} 1}{2} = \frac{\sqrt{2} + \log(1+\sqrt{2})}{2},$$ and the rest is just bookkeeping of the other constants and symmetries.
