transform of integrals
If differentiating in time multiplies the transform by s, then integrating in time ought to do the opposite — and it does. This rule is the mirror image of the derivative rule, and it is what lets you handle accumulated quantities: total charge that has flowed, total displacement, anything built up by integrating a rate.
The statement is: if L{f(t)} = F(s), then L{ integral from 0 to t of f(tau) d tau } = F(s)/s. Integrating from 0 to t in the time domain corresponds to dividing the transform by s. The reason is exactly the derivative rule run in reverse: the running integral g(t) satisfies g'(t) = f(t) with g(0) = 0, so s G(s) - 0 = F(s), giving G(s) = F(s)/s. Each extra integration divides by another factor of s.
In circuit analysis this is the transform that handles a capacitor, whose voltage is the integral of the current, and in mechanics it handles position as the integral of velocity. It also explains why the step function 1/s and the ramp 1/s^2 sit one factor of s apart: integrating the constant 1 gives the ramp t, and dividing 1/s by s gives 1/s^2. The clean symmetry between dividing by s (integrate) and multiplying by s (differentiate) is the algebraic heart of the whole transform calculus.
Since L{sin(omega t)} = omega/(s^2 + omega^2), the integral from 0 to t of sin(omega tau) d tau has transform omega/(s(s^2 + omega^2)), matching its known value (1 - cos(omega t))/omega.
Dividing the sine transform by s reproduces the transform of its running integral — a quick consistency check.
The rule assumes the integral starts at 0; an integral with a different lower limit carries an extra constant, so rewrite it as a running integral from 0 plus that constant before dividing by s.