Answer
$$\sec x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sin x}}{{{{\cos }^2}x}}} dx \cr
& {\text{write the denominator as a product}} \cr
& \int {\frac{{\sin x}}{{{{\cos }^2}x}}} dx = \int {\left( {\frac{{\sin x}}{{\cos x}}} \right)\left( {\frac{1}{{\cos x}}} \right)} dx \cr
& {\text{basic trigonometric identities }}\tan x = \frac{{\sin x}}{{\cos x}}{\text{ and }}\sec x = \frac{1}{{\cos x}} \cr
& = \int {\tan x\sec x} dx \cr
& {\text{use integration formula from table 4}}{\text{.2}}{\text{.1}} \cr
& = \sec x + C \cr} $$
