$\text{Let}f\left(t\right)\text{be defined for all positive}t\text{. The Laplace transform of}f\left(t\right)=\sin 2t·\sin 4t\text{i}$

The correct answer is:
$$\frac{16s}{(s^2+4)(s^2+36)}$$

Step 1: Simplify the trigonometric product
We begin with the given function:
$$f\left(t\right)=\sin 2t·\sin 4t$$
To rewrite the product as a sum or difference, we use the product-to-sum trigonometric identity:
$$\sin A·\sin B=\frac{1}{2}\left[\cos \right(A-B)-\cos (A+B\left)\right]$$
Substituting $A=4t$ and $B=2t$:
$$f\left(t\right)=\frac{1}{2}\left[\cos \right(4t-2t)-\cos (4t+2t\left)\right]$$
$$f\left(t\right)=\frac{1}{2}[\cos 2t-\cos 6t]$$

Step 2: Apply the Laplace transform
Using the linearity property of the Laplace transform:
$$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{2}\left[\mathcal{L}\right\{\cos 2t\}-\mathcal{L}\{\cos 6t\left\}\right]$$
Recall the standard Laplace transform formula for cosine:
$$\mathcal{L}\left\{\cos at\right\}=\frac{s}{s^2+a^2}$$
Applying this formula, we find:
$$\mathcal{L}\left\{\cos 2t\right\}=\frac{s}{s^2+2^2}=\frac{s}{s^2+4}$$
and
$$\mathcal{L}\left\{\cos 6t\right\}=\frac{s}{s^2+6^2}=\frac{s}{s^2+36}$$

Step 3: Combine and simplify the expression
Substitute the individual Laplace transforms back:
$$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{1}{2}$$
Factoring out $s$ and finding a common denominator:
$$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{s}{2}\left[\frac{(s^2+36)-(s^2+4)}{(s^2+4)(s^2+36)}\right]$$
$$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{s}{2}\left[\frac{32}{(s^2+4)(s^2+36)}\right]$$
Simplifying the coefficients gives the final result:
$$\mathcal{L}\left\{f\right(t\left)\right\}=\frac{16s}{(s^2+4)(s^2+36)}$$