# Integral of $\sec x$

 

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 $u = \sin x$ $\boxed { \int \sec x \; dx = \dfrac{1}{2} \ln \dfrac{|\sin x+1|}{|\sin x-1|} +c }$