Answer
$$ - {\operatorname{csch} ^{ - 1}}\left| {\sin x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\cos x}}{{\sin x\sqrt {{{\sin }^2}x + 1} }}dx} \cr
& {\text{substitute }}u = \sin x, \cr
& du = \cos xdx \cr
& \int {\frac{{\cos x}}{{\sin x\sqrt {{{\sin }^2}x + 1} }}dx} = \int {\frac{{du}}{{u\sqrt {{u^2} + 1} }}} \cr
& {\text{find the antiderivative using the theorem 6}}{\text{.8}}{\text{.6 }}\left( {{\text{see page 480}}} \right) \cr
& = - {\operatorname{csch} ^{ - 1}}\left| u \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \sin x, \cr
& = - {\operatorname{csch} ^{ - 1}}\left| {\sin x} \right| + C \cr} $$
