# Prove that \int \sec(x)=\ln(\sec x+\tan x).

## Question:

Prove that {eq}\displaystyle \int \sec(x)=\ln(\sec x+\tan x). {/eq}

## Integration using Trigonometric substitution:

Integration of a problem becomes simple if it is performed using trigonometric substitution. Here, trigonometric function is replaced by simple variables such as u,x etc. It is generally used when the conventional methods are unable to solve the integral problem. Some formulas that might be beneficial in solving the problem are:

{eq}1. \displaystyle \int {{z^n}} dz = \frac{{{z^{n + 1}}}}{{n + 1}} + C\\ 2. \displaystyle \int {\sin (z)dz = - \cos (z) + C} \\ 3. \displaystyle \int {\frac{1}{{({z^2} + 1)}}dz = arc(tan(z)) + C} \\ 4. \displaystyle \int {\frac{1}{z}} dz = \ln \left| z \right| + C {/eq}

where {eq}C {/eq} is an integration constant.

Given integral is

{eq}\int {\sec (x)dx} {/eq}

Multiplying numerical and denominator by {eq}\displaystyle {\sec (x) + \tan (x)} {/eq}

{eq}\displaystyle = \int {\sec (x)\frac{{\sec (x) + \tan (x)}}{{\sec (x) + \tan (x)}}dx} \\ \displaystyle = \int {\frac{{\sec (x)tan(x) + se{c^2}(x)}}{{\sec (x) + \tan (x)}}dx} {/eq}

Substituting {eq}\displaystyle u = \sec (x) + \tan (x) \to \frac{{du}}{{dx}} = \sec (x)\tan (x) + {\tan ^2}x \to dx = \frac{{du}}{{\sec (x)\tan (x) + {{\tan }^2}x}}\; \; \; \cdots \cdots (1) {/eq}, we get

{eq}\displaystyle = \int {\frac{1}{u}du} {/eq}

Using integral formula {eq}\displaystyle \int {\frac{1}{z}} dz = \ln \left| z \right| + C {/eq} we get

{eq}\displaystyle = \ln (u) {/eq}

Undo substitution from equation (1)

{eq}\displaystyle = \ln (\sec (x) + \tan (x))+C {/eq}

where {eq}C {/eq} is an integration constant. 