Finding the laplace transform of a squared function

Question

Find the laplace transform of

$f(t) = t^2cos^2(t)$.

I'm a bit stuck on this one. I know how to do this function if both components aren't squared due to using the laplace transform table.

Could someone give me a hint to start off this question and put me in the right direction? Thanks in advance.

Answer 1

Hint:

$$\mathcal{L}(t^2*f(t))=+\dfrac{{d}^2F(s)}{{ds}^2}$$

And you can rewrite the equation to the following: $${cos}^2t=\dfrac{1}{2}(cos(2t)+1)$$

Applying this gives:

$$\mathcal{L}(t^2*{cos}^2t)=\mathcal{L}(t^2*\dfrac{1}{2}(cos2t+1))$$ $$=\dfrac{1}{2}\mathcal{L}(t^2cos2t)+\dfrac{1}{2}\mathcal{L}(t^2)$$ This gives: $$\dfrac{1}{2}*\dfrac{d^2(\frac{s}{s^2+4})}{{ds}^2}+\dfrac{1}{s^3}=\dfrac{s(s^2-12)}{(s^2+4)^3}+\dfrac{1}{s^3}$$

Answer 2

Notice, by the definition of the Laplace transform:

$$\mathcal{L}_t\left[t^2\cos^2(t)\right]_{(s)}=\int_0^\infty t^2\cos^2(t)e^{-st}\space\text{d}t$$

---

Now, for the integral:

$$\int_0^\infty t^2\cos^2(t)e^{-st}\space\text{d}t=\frac{1}{2}\left[\int_0^\infty t^2\cos(2t)e^{-st}\space\text{d}t+\int_0^\infty t^2e^{-st}\space\text{d}t\right]$$

Use integration by parts:

• $$\int_0^\infty t^2\cos(2t)e^{-st}\space\text{d}t=$$ $$\left[\frac{t^2e^{-st}(2\sin(2t)-s\cos(2t))}{4+s^2}\right]_0^\infty-\frac{2}{s^2+4}\int_0^\infty te^{-st}(2\sin(t)-s\cos(2t))\space\text{d}t$$
• $$\int_0^\infty t^2e^{-st}\space\text{d}t=\left[-\frac{t^2e^{-st}}{s}\right]_0^\infty+\frac{2}{s}\int_0^\infty te^{-st}\space\text{d}t$$

You can also use a tool like WolframAlpha:

$$\text{LaplaceTransform[t^2 Cos[t]^2, t, s]}=\frac{2(s^6+24s^2+32)}{s^3(4+s^2)^3}$$