Laplace transforms of the functions tu(t) and u(t)sin(t) are respectively

The correct answer is A.

The Laplace transform of a function $f(t)$ is defined as

$$Lf(t): s = \int_0^\infty f(t) e^{-st} dt$$

The Laplace transform of $tu(t)$ is

$$Ltu(t): s = \int_0^\infty t e^{-st} dt = \frac{1}{s^2}$$

The Laplace transform of $u(t)sin(t)$ is

$$Lu(t)sin(t): s = \int_0^\infty u(t)sin(t) e^{-st} dt = \frac{s}{s^2 + 1}$$

Here is a brief explanation of each option:

• Option A is correct because it is the Laplace transform of $tu(t)$.
• Option B is incorrect because it is the Laplace transform of $u(t)$.
• Option C is incorrect because it is the Laplace transform of $u(t)cos(t)$.
• Option D is incorrect because it is the Laplace transform of $s$.