Answer
$$\int^{2\pi}_0\frac{\cos z}{\sqrt{4+3\sin z}}dz=\int^{\pi}_{-\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz=0$$
Work Step by Step
Part a) $$A=\int^{2\pi}_0\frac{\cos z}{\sqrt{4+3\sin z}}dz$$
We set $u=\sqrt{4+3\sin z}$, which means $$du=\frac{(4+3\sin z)'}{2\sqrt{4+3\sin z}}dz=\frac{3\cos z}{2\sqrt{4+3\sin z}}dz$$ $$\frac{\cos z}{\sqrt{4+3\sin z}}dz=\frac{2}{3}du$$
For $z=0$, we have $u=\sqrt{4+3\sin0}=\sqrt4=2$
For $z=2\pi$, we have $u=\sqrt{4+3\sin2\pi}=\sqrt4=2$
Therefore, $$A=\frac{2}{3}\int^2_2dz=0$$
Part b) $$A=\int^{\pi}_{-\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz$$
We set up the same substitution as in part a).
For $z=\pi$, we have $u=\sqrt{4+3\sin\pi}=\sqrt{4}=2$
For $z=-\pi$, we have $u=\sqrt{4+3\sin(-\pi)}=\sqrt4=2$
Therefore, $$A=\frac{2}{3}\int^2_2dz=0$$
