# Evaluate the integral \int \sec x \, dx using the substitution x = 2 \tan^{-1}t

## Question:

Evaluate the integral {eq}\int \sec x \, dx {/eq} using the substitution {eq}x = 2 \tan^{-1}t {/eq}

## Integral Substitution:

Indefinite integrals is the method of substitution,where one substitutes a variable for part of the integral, integrates the function with the new variable and then plugs the original value in place of the variable.

Formula used is given below: {eq}x=2\tan^{-1}(t)=\cos^{-1}\dfrac{1-t^{2}}{1+t^{2}} {/eq}

## Answer and Explanation:

{eq}\displaystyle\int sec xdx\\ x=2\tan^{-1}(t)\\ t=2\tan x {/eq}

Differentiate with respect to x, we get

{eq}dx = \frac{2}{1+t^2}\ dt {/eq}

Also,

{eq}\cos x = \dfrac{1-t^{2}}{1+t^{2}}\\ \Rightarrow \sec x= \dfrac{1+ t^{2}}{1- t^{2}} {/eq}

So, {eq}\int \sec x\ dx = \int \dfrac{1+ t^{2}}{1- t^{2}} \times \frac{2}{1+t^2}\ dt\\ = \int \dfrac{2}{1- t^{2}}\ dt\\ = \int \left( \dfrac{1}{1+ t} - \dfrac{1}{1- t} \right)\ dt\\ = \ln | 1+t| - \ln| 1-t| + C\\ = \ln \left( \frac{1+t}{1-t} \right) + C \\ = \ln \left( \frac{1+\tan (x/2)}{1-\tan(x/2)} \right) + C {/eq}

Where C is the constant of integration.

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from JEE (Main): Study Guide & Test Prep

Chapter 12 / Lesson 2