Question

Evaluate $\int {\dfrac{{\sec x + \tan x}}{{\sec x - \tan x}}dx = }$
A. $2\left( {\tan x + \sec x} \right) - x + c$
B. $\tan x - \sec x + x + c$
C. $2\left( {\tan x + \sec x} \right) + c$
D. $2\left( {\tan x + \sec x} \right) + x + c$

Answer 1

Here, we will first multiply the conjugate of the denominator to both the numerator and denominator of the given fraction. We will then use the algebraic identity and trigonometric identity to simplify the integrand. Then by using the suitable formula of integration, we will integrate the function.

Formula Used:
We will use the following formulas:
1. Trigonometric Identity: ${\sec ^2}x - {\tan ^2}x = 1$
2. The difference between the square of the numbers is given by the algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$
3. The square of the sum of two numbers is given by an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
4. Integral Formula: $\int {{{\sec }^2}xdx} = \tan x$
$\int {\sec x\tan xdx} = \sec x$
$\int {dx} = x$

Complete Step by Step Solution:
We are given an integral function $\int {\dfrac{{\sec x + \tan x}}{{\sec x - \tan x}}dx}$
Let the given integral function be $I$.
$I = \int {\dfrac{{\sec x + \tan x}}{{\sec x - \tan x}}dx}$
Now, we will multiply the integrand with the conjugate in the numerator and in the denominator, we get
$\Rightarrow I = \int {\dfrac{{\sec x + \tan x}}{{\sec x - \tan x}} \times \dfrac{{\sec x + \tan x}}{{\sec x + \tan x}}dx}$
Now, by using an algebraic identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$ in the denominator, we get
$\Rightarrow I = \int {\dfrac{{{{\left( {\sec x + \tan x} \right)}^2}}}{{{{\sec }^2}x - {{\tan }^2}x}}dx}$
By using the Trigonometric Identity ${\sec ^2}x - {\tan ^2}x = 1$, we get
$\Rightarrow I = \int {\dfrac{{{{\left( {\sec x + \tan x} \right)}^2}}}{1}dx}$
$\Rightarrow I = \int {{{\left( {\sec x + \tan x} \right)}^2}dx}$
Now, by using an algebraic identity ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we get
$\Rightarrow I = \int {\left( {{{\sec }^2}x + {{\tan }^2}x + 2\sec x\tan x} \right)dx}$
Trigonometric Identity:
Again using the Trigonometric Identity ${\sec ^2}x - {\tan ^2}x = 1$, we get
$\Rightarrow I = \int {\left( {{{\sec }^2}x + {{\sec }^2}x - 1 + 2\sec x\tan x} \right)dx}$
Adding the like terms, we get
$\Rightarrow I = \int {\left( {2{{\sec }^2}x - 1 + 2\sec x\tan x} \right)dx}$
Now, the integral sign distributes across the summation, we get
$\Rightarrow I = \int {2{{\sec }^2}xdx} - \int {dx} + \int {2\sec x\tan xdx}$
$\Rightarrow I = 2\int {{{\sec }^2}xdx} - \int {dx} + 2\int {\sec x\tan xdx}$
Now, by using the Integral formulas $\int {{{\sec }^2}xdx} = \tan x$, $\int {\sec x\tan xdx} = \sec x$ and $\int {dx} = x$ , we get
$\Rightarrow I = 2\tan x - x + 2\sec x + c$
Now, by taking out the common terms, we get
$\Rightarrow I = 2\left( {\tan x + \sec x} \right) - x + c$
Therefore, the value of $\int {\dfrac{{\sec x + \tan x}}{{\sec x - \tan x}}dx}$ is $2\left( {\tan x + \sec x} \right) - x + c$.

Thus, option (A) is the correct answer.

Note:
We know that Integration is the process of adding small parts to find the whole parts. Trigonometric identity is an equation which is always true for all the variables. Conjugate is a term where the sign is changed between two terms. When the integrand is in trigonometric function, then it satisfies the basic properties of integration. The given integral function is an indefinite integral since there is no limit in the integral. Whenever the integration is done with no limits, then an Arbitrary constant should be added.