Answer
$$\ln \left| {\sin x + \sqrt {\sin {x^2} + 1} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos x}}{{\sqrt {{{\sin }^2}x + 1} }}} dx \cr
& {\text{Let }}u = \sin x.{\text{ Then }}du = \cos xdx,{\text{ }}dx = \frac{1}{{\cos x}}du \cr
& {\text{Substituting you have}} \cr
& \int {\frac{{\cos x}}{{\sqrt {{{\sin }^2}x + 1} }}} dx = \int {\frac{{\cos x}}{{\sqrt {{u^2} + 1} }}} \left( {\frac{1}{{\cos x}}} \right)du \cr
& = \int {\frac{1}{{\sqrt {{u^2} + 1} }}} du \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{1}{{\sqrt {{u^2} + {a^2}} }}} du = \ln \left| {u + \sqrt {{u^2} + {a^2}} } \right| + C \cr
& \int {\frac{1}{{\sqrt {{u^2} + 1} }}} du = \ln \left| {u + \sqrt {{u^2} + 1} } \right| + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = \sin x \cr
& \int {\frac{{\cos x}}{{\sqrt {{{\sin }^2}x + 1} }}} dx = \ln \left| {\sin x + \sqrt {\sin {x^2} + 1} } \right| + C \cr} $$
