What is the integral of sin t dt with the interval x 0?

Let g(x) = interval [0, x] of sin t dt, and f(t) = sin t.

Since f(t) is a continuous function, the part one of the Fundamental Theorem of Calculus gives, g'(x) = sin x = f(x) (the original function).

If you are interested in the interval [x, 0] of sin t dt, then just put a minus sign in front of the integral and interchange places of 0 and x. So that,

g(x) = interval [x, 0] of sin t dt = -{ interval [0, x] of sin t dt}, then g'(x) = - sin x.