Answer
$$ - 4\sqrt {\cos x} + C $$
Work Step by Step
$$\eqalign{
& \int {2{{\left( {\cos x} \right)}^{ - 1/2}}\sin x} dx \cr
& {\text{use the substitution method}}{\text{:}} \cr
& u = \cos x,\,\,\,\,du = - \sin xdx,\,\,\,\,\,dx = \frac{{du}}{{ - \sin x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {2{{\left( {\cos x} \right)}^{ - 1/2}}\sin x} dx = \int {2{u^{ - 1/2}}\sin x} \left( {\frac{{du}}{{ - \sin x}}} \right) \cr
& = \int {2{u^{ - 1/2}}} \left( { - du} \right) \cr
& = - 2\int {{u^{ - 1/2}}} du \cr
& {\text{integrating by the power rule gives}} \cr
& = - 2\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr
& = - 4{u^{1/2}} + C \cr
& = - 4\sqrt u + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \cos x \cr
& = - 4\sqrt {\cos x} + C \cr} $$
