Let \(u = 1 + 3 \tan x\). Then, \(\frac{du}{dx} = 3 \sec^2 x\), so \(du = 3 \sec^2 x \, dx\).

Rearranging gives \(dx = \frac{du}{3 \sec^2 x}\).

Substitute into the integral:

\(\int \frac{\sqrt{u}}{\cos^2 x} \cdot \frac{du}{3 \sec^2 x} = \int \frac{\sqrt{u}}{3} \, du.\)

This simplifies to:

\(\frac{1}{3} \int u^{\frac{1}{2}} \, du.\)

Integrate:

\(\frac{1}{3} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{2}{9} u^{\frac{3}{2}}.\)

Evaluate from \(x = 0\) to \(x = \frac{\pi}{4}\):

When \(x = 0\), \(u = 1 + 3 \tan 0 = 1\).

When \(x = \frac{\pi}{4}\), \(u = 1 + 3 \tan \frac{\pi}{4} = 4\).

Substitute these limits into the integrated function:

\(\frac{2}{9} \left[ 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \right] = \frac{2}{9} (8 - 1) = \frac{2}{9} \times 7 = \frac{14}{9}.\)