Haqiqatan ham integralni baholashdan boshqa ko'p narsa qilish kerak emas.

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

E'tibor bering, teskari sinus va teskari kosinus hosilasi o'xshashligi sababli muqobil javob:

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