Find the Laplace transform, \(L\\{f(t)\\}=F(s)\), of (a) \(f(t)=2 \sin t+3 \cos 2 t\) (b) \(g(t)=\left[\left(1-e^{-t}\right) / t\right]\).

The given function is a linear combination of sine and cosine functions, so we can find the Laplace transforms of the individual functions and then sum them: \(F(s) = 2 L\\{\sin(t)\\} + 3 L\\{\cos(2t)\\}\) Applying the known Laplace transform formulas for sine and cosine functions: \(F(s) = 2 \frac{1}{s^2 + 1^2} + 3 \frac{2s}{s^2 + 2^2}\) Combining the terms, we have: \(F(s) = \frac{2}{s^2 + 1} + \frac{6s}{s^2 + 4}\) This is the Laplace transform of the function f(t).

The given function contains a term with a rational and an exponential term, so we need to find the Laplace transform as follows: \(G(s) = L\\{\frac{(1 - e^{-t})}{t}\\}\) To find the Laplace transform of this function, we will apply integration by parts: Let \(u = 1 - e^{-t}\) and \(dv = \frac{1}{t} dt\) Then we have \(du = e^{-t} dt\) and \(v = \ln(t)\). Applying integration to the function: \(G(s) = \int_0^{\infty} e^{-st} \frac{(1 - e^{-t})}{t} dt = \int_0^{\infty} vdu\) Expanding the integration using integration by parts, we get: \(G(s) = \lim_{a\to 0^+} [\ln(a)(1 - e^{-sa}) - \int_a^{\infty} \ln(t)e^{-st} dt-\int_0^{\infty} e^{-t}(\ln(t)e^{-st}dt]\) Now we have two integrals, we will apply integration by parts in the last integral term: Set \(u_1 = \ln(t)\) and \(dv_1 = -se^{-st} dt\) Then we have \(du_1 = \frac{1}{t} dt\) and \(v_1 = -\frac{e^{-st}}{s}\). So the integral term becomes: \(-\int_0^{\infty} e^{-t}\ln(t)e^{-st}dt = -\int_0^{\infty}u_1dv_1 = \{-\ln(t)\frac{e^{-st}}{s}\}\Bigg|_0^{\infty} + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt\) Substituting back the integrals into the expression of G(s), we have: \(G(s) = \lim_{a\to 0^+} [\ln(a)(1 - e^{-sa}) - \int_a^{\infty} \ln(t)e^{-st} dt + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt]\) As the limit of the remaining integral doesn't exist in elementary functions, we can represent G(s) with the remaining integral: \(G(s) = - \int_0^{\infty} \ln(t)e^{-st} dt + \frac{1}{s}\int_0^{\infty} \frac{e^{-st}}{t} dt\) This is the Laplace transform of the function g(t).