$\sin 2zx=0$
$2zx= n,\pi$
$x= \large\frac{n}{2}, $$ n \in o,1$
$\int \limits_0^{1/2} \sin 2zx dx + \int \limits_{1/2} ^1 -(\sin 2zx)dx$
$-\bigg[ \large\frac{\cos 2zx}{2z} \bigg]_0^{1/2}+ \bigg[ \large\frac{\cos 2 \pi x}{2x}\bigg]^1 _{1/2}$
$ -\large\frac{1}{2z}$$ [ \cos z - \cos 0 - \cos 2z+ \cos z ]$
$-\large\frac{1}{2z} $$[-2-1+ \cos z]$
$\qquad= \large\frac{-1}{z}$
Hence b is the correct answer.