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

More References and Links

1. Table of Integral Formulas
2. University Calculus - Early Transcendental - Joel Hass, Maurice D. Weir, George B. Thomas, Jr., Christopher Heil - ISBN-13 : 978-0134995540
   
3. Calculus - Gilbert Strang - MIT - ISBN-13 : 978-0961408824
   
4. Calculus - Early Transcendental - James Stewart - ISBN-13: 978-0-495-01166-8