1. **State the problem:** Find the Laplace Transform (L.T.) of the function $f(t) = t e^{2t} \sin t \cos 3t$. 2. **Recall the Laplace Transform definition:** The Laplace Transform of a function $f(t)$ is given by $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) \, dt$$ 3. **Simplify the function inside the transform:** Use the product-to-sum identity for trigonometric functions: $$\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$$ Applying this to $\sin t \cos 3t$: $$\sin t \cos 3t = \frac{1}{2}[\sin(4t) + \sin(-2t)] = \frac{1}{2}[\sin(4t) - \sin(2t)]$$ 4. **Rewrite the function:** $$f(t) = t e^{2t} \sin t \cos 3t = t e^{2t} \cdot \frac{1}{2}[\sin(4t) - \sin(2t)] = \frac{t e^{2t}}{2}[\sin(4t) - \sin(2t)]$$ 5. **Use linearity of Laplace Transform:** $$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( \mathcal{L}\{t e^{2t} \sin(4t)\} - \mathcal{L}\{t e^{2t} \sin(2t)\} \right)$$ 6. **Recall the Laplace Transform formula for $t e^{at} \sin(bt)$:** $$\mathcal{L}\{t e^{at} \sin(bt)\} = -\frac{d}{ds} \left( \frac{b}{(s - a)^2 + b^2} \right)$$ 7. **Calculate each term:** - For $\mathcal{L}\{t e^{2t} \sin(4t)\}$: $$F_1(s) = -\frac{d}{ds} \left( \frac{4}{(s - 2)^2 + 16} \right)$$ - For $\mathcal{L}\{t e^{2t} \sin(2t)\}$: $$F_2(s) = -\frac{d}{ds} \left( \frac{2}{(s - 2)^2 + 4} \right)$$ 8. **Compute derivatives:** Let $u = s - 2$. For $F_1(s)$: $$\frac{4}{u^2 + 16}$$ Derivative: $$\frac{d}{ds} \left( \frac{4}{u^2 + 16} \right) = \frac{d}{du} \left( \frac{4}{u^2 + 16} \right) \cdot \frac{du}{ds} = -\frac{8u}{(u^2 + 16)^2}$$ So, $$F_1(s) = -\left(-\frac{8u}{(u^2 + 16)^2}\right) = \frac{8(s - 2)}{((s - 2)^2 + 16)^2}$$ For $F_2(s)$: $$\frac{2}{u^2 + 4}$$ Derivative: $$\frac{d}{ds} \left( \frac{2}{u^2 + 4} \right) = -\frac{8u}{(u^2 + 4)^2}$$ So, $$F_2(s) = -\left(-\frac{8u}{(u^2 + 4)^2}\right) = \frac{8(s - 2)}{((s - 2)^2 + 4)^2}$$ 9. **Combine results:** $$\mathcal{L}\{f(t)\} = \frac{1}{2} \left( F_1(s) - F_2(s) \right) = \frac{1}{2} \left( \frac{8(s - 2)}{((s - 2)^2 + 16)^2} - \frac{8(s - 2)}{((s - 2)^2 + 4)^2} \right)$$ 10. **Simplify:** $$\mathcal{L}\{f(t)\} = 4(s - 2) \left( \frac{1}{((s - 2)^2 + 16)^2} - \frac{1}{((s - 2)^2 + 4)^2} \right)$$ **Final answer:** $$\boxed{\mathcal{L}\{t e^{2t} \sin t \cos 3t\} = 4(s - 2) \left( \frac{1}{((s - 2)^2 + 16)^2} - \frac{1}{((s - 2)^2 + 4)^2} \right)}$$