Find the arc length of the curve. How can I find the length of the loop in polar coordinates
$$r=\frac{a}{\cos^{4}( \frac{\varphi}{4})} $$ 
I do not even know where to start. Help please.

Parameterising
\begin{align}
\varphi(t) &= t \\
r(t) &= \frac{a}{\cos^{4}( \frac{t}{4})} 
\end{align}
and transforming to Cartesian coordinates
$$
x = r \cos \phi = \frac{a}{\cos^4(\frac{t}{4})} \cos t \quad\quad
y = r \sin \phi = \frac{a}{\cos^4(\frac{t}{4})} \sin t
$$
The image shows a plot for $a = 2$ and $t \in [-5,5]$.

$t=0$ is the point $(a, 0)$ we need to parameterise to the left crossing at $(x(t), 0)$ for $t = \pm \pi$.
The arc lenght is
$$
ds^2 = dx^2 + dy^2 \quad \Rightarrow \quad
ds = 
\sqrt{
\left(\frac{dx}{d\varphi}\right)^2 + 
\left(\frac{dy}{d\varphi}\right)^2} d\varphi
$$
and the derivatives are
$$
\frac{dx}{d\varphi} = \frac{dr}{d\varphi} \cos \varphi - r \sin \varphi
\quad\quad
\frac{dy}{d\varphi} = \frac{dr}{d\varphi} \sin \varphi + r \cos \varphi
$$
so we get the differential arc lenght in polar coordinates
$$
\left(\frac{ds}{d\varphi}\right)^2 =
\left(\frac{dr}{d\varphi}\right)^2 + r^2
$$
and using the parameterisation for $r$ in terms of $\varphi$:
$$
r'(\varphi) 
= -4 \frac{a}{\cos^5(\frac{t}{4})}
\left(-\sin\left(\frac{t}{4}\right) \frac{1}{4}\right)
= a \frac{\sin(\frac{t}{4})}{\cos^5(\frac{t}{4})} 
$$
Returning to the differential arc length element $ds$ we get
\begin{align}
s'(\varphi)^2 &= 
r'(\varphi)^2 + r^2 \\
&=
\left(\frac{a\sin(\frac{\varphi}{4})}{\cos^5(\frac{\varphi}{4})}\right)^2 +
\left(\frac{a}{\cos^{4}( \frac{\varphi}{4})}\right)^2 \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\left(\tan^2\left({\frac{\varphi}{4}}\right)+ 1\right) \\
&=
\frac{a^2}{\cos^8(\frac{\varphi}{4})}
\frac{1}{\cos^2\left({\frac{\varphi}{4}}\right)} \\
&=
\frac{a^2}{\cos^{10}(\frac{\varphi}{4})}
\end{align}
which turned out simpler than feared. So we can now give the arc length:
\begin{align}
s 
&= \int\limits_{-\pi}^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2\int\limits_0^\pi \frac{ds}{d\varphi} d\varphi \\
&= 2a \int\limits_0^\pi \frac{d\varphi}{\cos^{5}(\frac{\varphi}{4})} \\
&= 8a \int\limits_0^{\pi/4} \frac{du}{\cos^{5}(u)} \\
&= 8a (F(\pi/4) - F(0))
\end{align}
This has a longish anti-derivative which Maxima computes to
$$
F(\varphi) =
\frac{3\,\ln\left( \sin\left( u\right) +1\right) }{16}-\frac{3\,\ln\left( \sin\left( u\right) -1\right) }{16}-\frac{3\,{\sin\left( u\right) }^{3}-5\,\sin\left( u\right) }{8\,{\sin\left( u\right) }^{4}-16\,{\sin\left( u\right) }^{2}+8}
+ C
$$
and
\begin{align}
s 
&= 
a \left(
\frac{3\,\ln\left( \frac{1}{\sqrt{2}}+1\right)}{2}-
\frac{3\,\ln\left( 1-\frac{1}{\sqrt{2}}\right)}{2}+
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left( \frac{\sqrt{2}+1}{\sqrt{2}-1} \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
\frac{3}{2}
\ln\left((\sqrt{2}+1)^2 \right) +
7\,\sqrt{2}
\right) \\
&=
a \left(
3 \ln\left( \sqrt{2}+1 \right) +
7\,\sqrt{2}
\right) \\
&= 12.54361569767029 \, a
\end{align}
