Question

Find $\int {\dfrac{{\sec x}}{{\sec x + \tan x}}dx} $ equals:
(A) $\tan x - \sec x + c$
(B) $\log \left( {1 + \sec x} \right) + c$
(C) $\sec x + \tan x + c$
(D) $\log \sin x + \log \cos x + c$

Answer 1

Hint: The given question requires us to integrate a trigonometric rational function of x with respect to x. We evaluate the given integral using the substitution method. We substitute the denominator of the rational function as a new variable and then convert the integral in respect to the new variable.

Complete step-by-step answer:
The given question requires us to evaluate the integral $\int {\dfrac{{\sec x}}{{\sec x + \tan x}}dx} $ in variable x.
So, we have, $\int {\dfrac{{\sec x}}{{\sec x + \tan x}}dx} $
But, it is very difficult to integrate the function directly. So, we can assign a new variable to the denominator of the rational function.
So, let $\sec x + \tan x = t$.
Then, differentiating both sides of the equation, we get,
$ \Rightarrow \left( {\sec x\tan x + {{\sec }^2}x} \right)dx = dt$
Factoring out secant of x from the left side of equation, we get,
$ \Rightarrow \sec x\left( {\tan x + \sec x} \right)dx = dt$
Finding the value of $dx$ from the equation, we get,
$ \Rightarrow dx = \dfrac{{dt}}{{\sec x\left( {\tan x + \sec x} \right)}}$
So, the integral $\int {\dfrac{{\sec x}}{{\sec x + \tan x}}dx} $ can be simplified by substituting the value of $\left( {dx} \right)$ in terms of $\left( {dt} \right)$ as obtained above. So, we get,
\[\int {\dfrac{{\sec x}}{{\sec x + \tan x}}dx} = \int {\dfrac{{\sec x}}{{\sec x + \tan x}} \times \dfrac{{dt}}{{\sec x\left( {\tan x + \sec x} \right)}}} \]
Cancelling the common factors in numerator and denominator, we get,
\[ \Rightarrow \int {\dfrac{{dt}}{{{{\left( {\sec x + \tan x} \right)}^2}}}} \]
Substituting the term \[\left( {\sec x + \tan x} \right)\] as t.
\[ \Rightarrow \int {\dfrac{{dt}}{{{t^2}}}} \]
Now, using the power rule of integration $\int {{x^n}dx} = \left( {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right) + c$, we get,
\[ \Rightarrow - \dfrac{1}{t} + c\]
Substituting back the value of t, we get,
\[ \Rightarrow - \left( {\dfrac{1}{{\sec x + \tan x}}} \right) + c\]
Now, multiplying the numerator and denominator by $\sec x - \tan x$, we get,
 \[ \Rightarrow - \left( {\dfrac{1}{{\sec x + \tan x}}} \right)\left( {\dfrac{{\sec x - \tan x}}{{\sec x - \tan x}}} \right) + c\]
Using the algebraic identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$,
\[ \Rightarrow - \left( {\dfrac{{\sec x - \tan x}}{{{{\sec }^2}x - {{\tan }^2}x}}} \right) + c\]
Now, we use the trigonometric identity \[{\sec ^2}x - {\tan ^2}x = 1\]. So, we get,
\[ \Rightarrow - \left( {\sec x - \tan x} \right) + c\]
Opening the brackets and simplifying the expression, we get,
\[ \Rightarrow \tan x - \sec x + c\]
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: The indefinite integrals of certain functions may have more than one answer in different forms. However, all these forms are correct and interchangeable into one another. Indefinite integral gives us the family of curves as we don’t know the exact value of the arbitrary constant. We must know the method of substitution in order to solve this integral. One must take care of the calculations in order to get to the final answer.