convolution theorem
Multiplication is easy in the s-domain but means something subtle back in time. The convolution theorem answers the question: if you multiply two transforms together, what time function does the product correspond to? The answer is not the product of the two time functions — it is their convolution, a blending operation that mixes one function's past into the other.
The convolution of f and g is written (f * g)(t) = integral from 0 to t of f(tau) g(t - tau) d tau: you flip g, slide it, and accumulate the overlap. The theorem states that L{f * g} = F(s) G(s). So a product in the s-domain is a convolution in time, and conversely the inverse transform of a product F(s) G(s) is the convolution of the two inverses. This is what lets you invert a product when you recognize each factor separately but cannot find the product in any table.
Physically the convolution integral is how a linear time-invariant system responds to an arbitrary input: you treat the input as a continuous stream of impulses, weight each by the system's impulse response, and add up the echoes. That is exactly the meaning of (input * impulse response). The convolution theorem is therefore the bridge between the transfer function in the s-domain (a simple product H(s) X(s)) and the time-domain superposition integral — the same fact wearing two costumes.
The inverse of 1/(s(s^2 + 1)) is the convolution of 1 (inverse of 1/s) with sin(t) (inverse of 1/(s^2 + 1)): integral from 0 to t of sin(tau) d tau = 1 - cos(t).
When the product is not in the table, convolving the two known inverses delivers the answer.
The single most common error is to assume L{f g} equals F(s) G(s); it does not — the product of transforms equals the transform of the convolution, never the transform of the pointwise product.