Answer
$a.\quad 0$
$ b.\quad 0$
Work Step by Step
$a.$
Applying th.7, let $u=g(z)=4+3\sin z$ (a continuous function).
Then, $\quad du=g'(z)dz=\cos z\cdot 3\cdot dz=3\cos zdz$
and, the borders change to $g(0)=4+0=4$ and $g(2\pi)=4+0=4.$
$\displaystyle \int_{0}^{2\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz=\int_{4}^{4}u^{-1/2}\cdot\frac{1}{3}du =0$
(zero width integral)
$b.$
Applying the same substitution,
the borders here change to $g(-\pi)=4+0=4$ and $g(\pi)=4+0=4$.
Again, the integral will have zero width,
$\displaystyle \int_{-\pi}^{\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz=\int_{4}^{4}u^{-1/2}\cdot\frac{1}{3}du =0$
