1. **Problem statement:** Find the function $y = \int_0^x \tan t \, dt$ for $0 \leq x \leq \frac{\pi}{6}$. 2. **Formula and rules:** The integral of $\tan t$ can be found using the identity $\tan t = \frac{\sin t}{\cos t}$. The integral formula is $$\int \tan t \, dt = -\ln|\cos t| + C$$ where $C$ is the constant of integration. 3. **Evaluate the definite integral:** $$y = \int_0^x \tan t \, dt = [-\ln|\cos t|]_0^x = -\ln|\cos x| + \ln|\cos 0|$$ Since $\cos 0 = 1$, this simplifies to $$y = -\ln|\cos x| + 0 = -\ln(\cos x)$$ 4. **Final answer:** $$\boxed{y = -\ln(\cos x) \text{ for } 0 \leq x \leq \frac{\pi}{6}}$$ This function starts at $y=0$ when $x=0$ and increases as $x$ approaches $\frac{\pi}{6}$ because $\cos x$ decreases, making $-\ln(\cos x)$ increase.