Step 1: Rewrite the integrand in a usable form.
Since $\frac{1}{\cos^4x}=\sec^4x$, we are integrating $\sec^4x$ from $0$ to $\frac{\pi}{4}$.
Step 2: Split off one $\sec^2x$.
Write $\sec^4x=\sec^2x\cdot\sec^2x$, and replace one factor using $\sec^2x=1+\tan^2x$.
Step 3: Substitute the tangent.
Let $t=\tan x$, so $dt=\sec^2x\,dx$. The remaining $\sec^2x$ becomes $1+t^2$.
Step 4: Change the limits.
When $x=0$, $t=\tan 0=0$; when $x=\frac{\pi}{4}$, $t=\tan\frac{\pi}{4}=1$. So the integral becomes $\int_0^1 (1+t^2)\,dt$.
Step 5: Integrate the polynomial.
$\int_0^1 (1+t^2)\,dt=\left[t+\frac{t^3}{3}\right]_0^1$.
Step 6: Evaluate at the limits.
Plugging in gives $1+\frac{1}{3}=\frac{4}{3}$, which is option 2.
\[ \boxed{\frac{4}{3}} \]