1. **Problem (a):** Evaluate the integral $$\int_0^1 \frac{\ln(x)}{\sqrt{x}} \, dx$$. 2. Let us use the substitution $$x = t^2$$, then $$dx = 2t \, dt$$ and $$\sqrt{x} = t$$. Also, when $$x=0$$, $$t=0$$; when $$x=1$$, $$t=1$$. 3. Rewriting the integral: $$\int_0^1 \frac{\ln(x)}{\sqrt{x}} \, dx = \int_0^1 \frac{\ln(t^2)}{t} (2t) \, dt = \int_0^1 2 \ln(t^2) \, dt = \int_0^1 4 \ln(t) \, dt$$ 4. Now evaluate $$\int_0^1 4 \ln(t) \, dt$$. Recall the formula $$\int_0^1 \ln(t) \, dt = -1$$. 5. Therefore: $$\int_0^1 4 \ln(t) \, dt = 4 \times (-1) = -4$$ 6. So, the answer for (a) is $$\boxed{-4}$$. --- 7. **Problem (b):** Evaluate the integral $$\int_0^{\pi/2} \sec(x) \, dx$$. 8. Recall the integral formula: $$\int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C$$ 9. Apply the definite integral: $$\int_0^{\pi/2} \sec(x) \, dx = \left[ \ln|\sec(x) + \tan(x)| \right]_0^{\pi/2}$$ 10. Evaluate the limit as $$x \to \pi/2^-$$ for $$\sec(x) + \tan(x)$$: Both $$\sec(x)$$ and $$\tan(x)$$ approach $$+\infty$$, so the expression tends to infinity. 11. This means the integral diverges to infinity. 12. Therefore, the integral $$\int_0^{\pi/2} \sec(x) \, dx$$ is divergent (does not converge to a finite value). --- **Final answers:** (a) $$\boxed{-4}$$ (b) The integral $$\int_0^{\pi/2} \sec(x) \, dx$$ diverges and does not have a finite value.