# calcul – How to calculate \$ int_ {0} ^ { pi} | sin (x) | dx \$ from his indefinite integral

I have trouble understanding why, when I try to calculate $$int_ {0} ^ { pi} | sin (x) | dx$$ of its indefinite completeness I seem to get the wrong result of $$0$$.

Of course $$| sin (x) | = sin (x)$$ sure $$(0, pi)$$ so I know that the result should be $$2$$. But considering the indefinite integral

$$int | sin (x) | = – cos (x) sgn ( sin (x))$$

($$sgn$$ is the function of the sign) I use fallacious reasoning to conclude that

$$int_ {0} ^ { pi} | sin (x) | dx = (- cos (x) sgn ( sin (x))) _ {0} ^ { pi} = – (- 1) (0) – (- 1) (0)) = 0$$

I think the problem is how do I use the signaling function, but as far as I know $$sgn ( sin (0)) = sgn ( sin ( pi)) = 0$$ and $$| sin (x) |$$ is continuous on $$(0, pi)$$. What am I doing wrong here?