Answer
$$\frac{{\sin x}}{2}\sqrt {4 + {{\sin }^2}x} + \frac{4}{2}\ln \left( {\sin x + \sqrt {4 + {{\sin }^2}x} } \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\cos x\sqrt {4 + {{\sin }^2}x} } dx \cr
& {\text{Let }}u = \sin x,{\text{ }}du = \cos xdx \cr
& \int {\cos x\sqrt {4 + {{\sin }^2}x} } dx = \int {\sqrt {{{\left( 2 \right)}^2} + {u^2}} } du \cr
& {\text{Use the Table of Integrals on the Reference Pages }}\left( {{\text{510}}} \right) \cr
& *\int {\sqrt {{a^2} + {u^2}} du} = \frac{u}{2}\sqrt {{a^2} + {u^2}} + \frac{{{a^2}}}{2}\ln \left( {u + \sqrt {{a^2} + {u^2}} } \right) + C,{\text{ so}} \cr
& \int {\sqrt {{{\left( 2 \right)}^2} + {u^2}} } du = \frac{u}{2}\sqrt {{{\left( 2 \right)}^2} + {u^2}} + \frac{4}{2}\ln \left( {u + \sqrt {{{\left( 2 \right)}^2} + {u^2}} } \right) + C \cr
& = \frac{u}{2}\sqrt {4 + {u^2}} + \frac{4}{2}\ln \left( {u + \sqrt {4 + {u^2}} } \right) + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{{\sin x}}{2}\sqrt {4 + {{\sin }^2}x} + \frac{4}{2}\ln \left( {\sin x + \sqrt {4 + {{\sin }^2}x} } \right) + C \cr} $$
