Evaluate the integral
\[ \int \sec x \; dx \]
Use \( \sec x = \dfrac{1}{\cos x} \) to rewrite the integral as
\[ \int \sec x \; dx = \int \dfrac{1}{\cos x} \; dx \]
Multiply both numerator and denominator of the intrgrand on the right by \( \cos x \)
\[ \int \sec x \; dx = \int \dfrac{\cos x}{\cos^2 x} \; dx \]
Use trigonometric identity \( \cos^2 x = 1 - \sin^2 x \) on the right and write
\[ \int \sec x \; dx = \int \dfrac{\cos x}{1 - \sin^2 x} \; dx \]
Use Integration by Substitution: \( u = \sin x \) so that \( du = \cos x dx \), and the given integral can be written as
\[ \int \sec x \; dx = \int \dfrac{1}{1 - u^2} \; du \]
Use partial fractions decomposition to write the integrand as
\[ \dfrac{1}{ (1-u^2)} = \dfrac{1}{2\left(u+1\right)}-\dfrac{1}{2\left(u-1\right)} \]
Substitute in the integral
\[ \int \sec x \; dx = \dfrac{1}{2} \int \dfrac{1}{u+1} du - \dfrac{1}{2} \int \dfrac{1}{u-1} du \]
Use the formula of integration \( \displaystyle\int \dfrac{f'(x)}{f(x)} \; dx = \ln | f(x)| +c \) to obtain
\[ \int \sec x \; dx = \dfrac{1}{2} \ln |u+1| - \dfrac{1}{2} \ln |u-1| + c\]
Group the logarithmic expressions using the properties \( \quad \ln \dfrac{a}{b } = \ln a - \ln b \) to write
\[ \int \sec x \; dx = \dfrac{1}{2} \ln \dfrac{|u+1|}{|u-1|} + c \]
Substitute back \( u = \sin x \)
\[ \boxed { \int \sec x \; dx = \dfrac{1}{2} \ln \dfrac{|\sin x+1|}{|\sin x-1|} +c } \]
