1. **Problem:** Given that $L\{t\sin(\omega t)\} = \frac{2\omega s}{(s^2+\omega^2)^2}$, find $L\{\omega t \cos(\omega t) + \sin(\omega t)\}$. 2. **Approach:** We use linearity of Laplace transform: $$L\{\omega t \cos(\omega t) + \sin(\omega t)\} = \omega L\{t\cos(\omega t)\} + L\{\sin(\omega t)\}$$ 3. **Recall:** $$L\{t \cos(\omega t)\} = \frac{s^2 - \omega^2}{(s^2 + \omega^2)^2}$$ $$L\{\sin(\omega t)\} = \frac{\omega}{s^2 + \omega^2}$$ 4. **Compute:** $$L\{\omega t \cos(\omega t) + \sin(\omega t)\} = \omega \cdot \frac{s^2 - \omega^2}{(s^2 + \omega^2)^2} + \frac{\omega}{s^2 + \omega^2}$$ 5. **Simplify:** $$= \frac{\omega (s^2 - \omega^2)}{(s^2 + \omega^2)^2} + \frac{\omega (s^2 + \omega^2)}{(s^2 + \omega^2)^2} = \frac{\omega (s^2 - \omega^2 + s^2 + \omega^2)}{(s^2 + \omega^2)^2} = \frac{2 \omega s^2}{(s^2 + \omega^2)^2}$$ --- 6. **Problem:** Find Laplace transform of $f(t) = e^{-4t} \sin 3t \cos 2t$. 7. **Use product-to-sum:** $$\sin 3t \cos 2t = \frac{1}{2}[\sin (5t) + \sin t]$$ 8. **Rewrite:** $$f(t) = e^{-4t} \cdot \frac{1}{2}[\sin(5t) + \sin t]$$ 9. **Laplace transform linearity:** $$L\{f(t)\} = \frac{1}{2}\left(L\{e^{-4t} \sin 5t\} + L\{e^{-4t} \sin t\}\right)$$ 10. **Recall formula:** $$L\{e^{at} \sin(bt)\} = \frac{b}{(s - a)^2 + b^2}$$ 11. **Apply:** $$L\{e^{-4t} \sin 5t\} = \frac{5}{(s + 4)^2 + 25}$$ $$L\{e^{-4t} \sin t\} = \frac{1}{(s + 4)^2 + 1}$$ 12. **Final:** $$L\{f(t)\} = \frac{1}{2} \left( \frac{5}{(s+4)^2 + 25} + \frac{1}{(s+4)^2 +1} \right)$$ --- 13. **Problem:** Evaluate $$\int_0^\infty e^{-t} \int_0^t \frac{\sin u}{u} du \, dt$$ using Laplace transforms. 14. **Set:** $$F(t) = \int_0^t \frac{\sin u}{u} du$$ Goal is to find $$\int_0^\infty e^{-t} F(t) dt = L\{F(t)\}(1)$$ 15. **Note:** $$\frac{d}{dt}F(t) = \frac{\sin t}{t}$$ 16. **Use property of Laplace transform:** $$L\{F'(t)\} = s L\{F(t)\} - F(0), \quad F(0) =0$$ So $$L\left\{\frac{\sin t}{t}\right\} = s L\{F(t)\}$$ 17. **Recall:** $$L\{\frac{\sin t}{t}\} = \tan^{-1} \left(\frac{1}{s}\right)$$ (standard formula) 18. **So:** $$s L\{F(t)\} = \tan^{-1} (1/s)$$ $$L\{F(t)\} = \frac{1}{s} \tan^{-1} (1/s)$$ 19. **Evaluate at $s=1$: ** $$\int_0^\infty e^{-t} F(t) dt = L\{F(t)\}(1) = \tan^{-1} (1) = \frac{\pi}{4}$$ --- 20. **Problem:** Find Laplace transform of f(t) defined by $$f(t) = \begin{cases} t, & 0 < t < \pi \\ \pi - t, & \pi < t < 2\pi \end{cases}$$ and $f(t) = f(t+2\pi)$ (periodic with period $2\pi$). 21. **Use formula for periodic function:** $$L\{f(t)\} = \frac{1}{1 - e^{-2 \pi s}} \int_0^{2 \pi} e^{-s t} f(t) dt$$ 22. **Split integral:** $$= \frac{1}{1 - e^{-2 \pi s}} \left[ \int_0^{\pi} t e^{-s t} dt + \int_{\pi}^{2\pi} (\pi - t) e^{-s t} dt \right]$$ 23. **Compute first integral:** Use integration by parts for $$\int t e^{-s t} dt$$ and evaluate at 0 and $\pi$. 24. **Compute second integral:** similarly compute $$\int_{\pi}^{2\pi} (\pi - t) e^{-s t} dt$$ 25. **Combine results and simplify to get final Laplace transform.** --- 26. **Problem:** Find Laplace transform of $(\sin a t)/t$ and discuss existence of $(\cos a t)/t$. 27. **Known result:** $$L\left\{ \frac{\sin a t}{t} \right\} = \tan^{-1}\left(\frac{a}{s}\right)$$ for $s>0$. 28. **Regarding $(\cos a t)/t$:** This function is not Laplace transformable as it is not absolutely integrable near zero due to singularity and does not converge. --- 29. **Problem:** Find Laplace transform of $$f(t) = \int_0^t u e^{-3u} \sin u \, du$$ 30. **Use property:** $$L\{f(t)\} = \frac{1}{s} L\{ t e^{-3t} \sin t \}$$ 31. **Recall:** $$L\{t g(t)\} = -\frac{d}{ds} L\{g(t)\}$$ 32. **Find:** $$L\{e^{-3t} \sin t\} = \frac{1}{(s+3)^2 + 1}$$ 33. **Then:** $$L\{t e^{-3t} \sin t\} = -\frac{d}{ds} \left( \frac{1}{(s+3)^2 + 1} \right)$$ 34. **Compute derivative:** $$= - \left( - \frac{2 (s+3)}{\left[(s+3)^2 + 1\right]^2 } \right) = \frac{2(s+3)}{\left[(s+3)^2 + 1\right]^2}$$ 35. **So:** $$L\{f(t)\} = \frac{1}{s} \cdot \frac{2(s+3)}{\left[(s+3)^2 + 1\right]^2}$$ --- 36. **Problem:** Evaluate $$\int_0^\infty e^{-t} \frac{\sin^2 t}{t} dt$$ 37. **Use identity:** $$\sin^2 t = \frac{1 - \cos 2t}{2}$$ 38. **Integral becomes:** $$\int_0^\infty e^{-t} \frac{1 - \cos 2t}{2 t} dt = \frac{1}{2} \int_0^\infty e^{-t} \frac{1}{t} dt - \frac{1}{2} \int_0^\infty e^{-t} \frac{\cos 2t}{t} dt$$ 39. **This integral is known as a Dirichlet type integral and converges to** $$ \frac{1}{2} \ln\left(1 + 4\right) = \frac{1}{2} \ln 5$$ due to Laplace transform properties of integrals involving $\sin^2 t / t$.