Find the Laplace of following $L(t^2 + \cos 2t \cos t + \sin ^2 2t)$

To find the Laplace transform of the given function $f(t) = t^2 + \cos(2t)\cos(t) + \sin^2(2t)$, we’ll break it down into individual terms and use the linearity property of the Laplace transform. The Laplace transform of a sum is the sum of the Laplace transforms of the individual terms. Here’s the step-by-step solution:

1. Laplace Transform of $t^2$: The Laplace transform of $t^n$ is $\dfrac{n!}{s^{n+1}}$, where $n$ is a non-negative integer. In this case, $n = 2$. $$\mathcal{L}\{t^2\} = \dfrac{2!}{s^{2+1}} = \dfrac{2}{s^3}$$

2. Laplace Transform of $\cos(2t)\cos(t)$: Use the trigonometric identity $\cos(A)\cos(B) = \dfrac{1}{2}[\cos(A + B) + \cos(A - B)]$. $$\cos(2t)\cos(t) = \dfrac{1}{2}[\cos(2t + t) + \cos(2t - t)] = \dfrac{1}{2}[\cos(3t) + \cos(t)]$$

   Now, we’ll take the Laplace transform of both terms: $$\mathcal{L}\left\{\dfrac{1}{2}[\cos(3t) + \cos(t)]\right\} = \dfrac{1}{2} \left(\mathcal{L}\{\cos(3t)\} + \mathcal{L}\{\cos(t)\}\right)$$

   The Laplace transform of $\cos(at)$ is $\dfrac{s}{s^2 + a^2}$, so: $$\mathcal{L}\{\cos(3t)\} = \dfrac{s}{s^2 + 3^2} = \dfrac{s}{s^2 + 9}$$ $$\mathcal{L}\{\cos(t)\} = \dfrac{s}{s^2 + 1^2} = \dfrac{s}{s^2 + 1}$$

   Putting it all together: $$\mathcal{L}\left\{\dfrac{1}{2}[\cos(3t) + \cos(t)]\right\} = \dfrac{1}{2} \left(\dfrac{s}{s^2 + 9} + \dfrac{s}{s^2 + 1}\right)$$

3. Laplace Transform of $\sin^2(2t)$: The Laplace transform of $\sin^2(at)$ can be found using a trigonometric identity and the Laplace transform of $\sin(at)$. The identity is $\sin^2(A) = \dfrac{1}{2}(1 - \cos(2A))$. $$\sin^2(2t) = \dfrac{1}{2}(1 - \cos(4t))$$

   Now, we’ll take the Laplace transform of both terms: $$\mathcal{L}\left\{\dfrac{1}{2}(1 - \cos(4t))\right\} = \dfrac{1}{2} \left(\mathcal{L}\{1\} - \mathcal{L}\{\cos(4t)\}\right)$$

   The Laplace transform of the constant function 1 is $\dfrac{1}{s}$. $$\mathcal{L}\{1\} = \dfrac{1}{s}$$

   The Laplace transform of $\cos(at)$ is $\dfrac{s}{s^2 + a^2}$, so: $$\mathcal{L}\{\cos(4t)\} = \dfrac{s}{s^2 + 4^2} = \dfrac{s}{s^2 + 16}$$

   Putting it all together: $$\mathcal{L}\left\{\dfrac{1}{2}(1 - \cos(4t))\right\} = \dfrac{1}{2} \left(\dfrac{1}{s} - \dfrac{s}{s^2 + 16}\right)$$

Now, we can add up all these Laplace transforms to find the Laplace transform of the original function $f(t)$:

$$\begin{aligned} \mathcal{L}\{f(t)\} &= \mathcal{L}\{t^2\} + \mathcal{L}\{\dfrac{1}{2}[\cos(3t) + \cos(t)]\} + \mathcal{L}\{\dfrac{1}{2}(1 - \cos(4t))\} \\ &= \dfrac{2}{s^3} + \dfrac{1}{2} \left(\dfrac{s}{s^2 + 9} + \dfrac{s}{s^2 + 1}\right) + \dfrac{1}{2} \left(\dfrac{1}{s} - \dfrac{s}{s^2 + 16}\right) \end{aligned}$$