There really isn’t too much to do other than to evaluate the integral.

\[\int{{6\cos \left( z \right) + \frac{4}{{\sqrt {1 - {z^2}} }}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6\sin \left( z \right) + 4{{\sin }^{ - 1}}\left( z \right) + c}}\]

Note that because of the similarity of the derivative of inverse sine and inverse cosine an alternate answer is,

\[\int{{6\cos \left( z \right) + \frac{4}{{\sqrt {1 - {z^2}} }}\,dz}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6\sin \left( z \right) - 4{{\cos }^{ - 1}}\left( z \right) + c}}\]