Question

Find the solution of the integral $\dfrac{{\tan x}}{{\left( {\sec x + \tan x} \right)}}dx$

Answer 1

Hint: We have to integrate indefinite integral $\dfrac{{\tan x}}{{\left( {\sec x + \tan x} \right)}}dx$ with respect to $x$. The process of differentiation and integration are inverses of each other. Hence, $\int {f'\left( x \right)} dx = f\left( x \right) + C$ where $C$ is the arbitrary constant called the constant of integration. We solve this function by rationalizing it and after rationalizing we will simplify it and substitute ${\sec ^2}x - {\tan ^2}x = 1$. We will split the single integral into multiple integrals. We also know various standard integration formulae; we will put the values of standard formulae and obtain the final answer in terms of $x$ only.
Property for splitting the functions:
For functions $f\left( x \right)$ and $g\left( x \right)$, $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]} dx = \int {f\left( x \right)} dx \pm \int {g\left( x \right)} $
I.e. The indefinite integral of the sum/difference is equal to the sum/difference of integrals.

Complete step by step solution:
Given $\int {\dfrac{{\tan x}}{{(\sec x + \tan x)}}} dx$
On rationalizing, we get
$ \Rightarrow \int {\dfrac{{\tan x}}{{(\sec x + \tan x)}} \times \dfrac{{(\sec x - \tan x)}}{{(\sec x - \tan x)}}} dx$
$ \Rightarrow \int {\dfrac{{\tan x(\sec x - \tan x)}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx$
Simplify by multiplying
$ \Rightarrow \int {\dfrac{{\tan x\sec x - {{\tan }^1}x{{\tan }^1}x}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx$
Use the power rule ${x^m} + {x^n} = {x^{m + n}}$ to combine the exponents.
$ \Rightarrow \int {\dfrac{{\tan x\sec x - {{\tan }^{1 + 1}}x}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx$
$ \Rightarrow \int {\dfrac{{\tan x\sec x - {{\tan }^2}x}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx$
We know that $1 + {\tan ^2}x = {\sec ^2}x$. Therefore, ${\sec ^2}x - {\tan ^2}x = 1$
$ \Rightarrow \int {\dfrac{{\tan x\sec x - {{\tan }^2}x}}{1}} dx$
$ \Rightarrow \int {\left( {\tan x\sec x - {{\tan }^2}x} \right)} dx$
Split the single integral into multiple integrals.
 $ \Rightarrow \int {\tan x\sec xdx - \int {{{\tan }^2}} } xdx$
Put ${\tan ^2}x = {\sec ^2}x + 1$.
$ \Rightarrow \int {\tan x\sec xdx} - \int {({{\sec }^2}x - 1)dx} $
Split the single integral into multiple integrals.
$ \Rightarrow \int {\tan x\sec x} dx - \int {{{\sec }^2}xdx + \int {1dx} } $
Since the derivative of $\sec x$ is $\sec x\tan x$, the integral of $\sec x\tan x$ is $\sec x$ and Integral of ${\sec ^2}x$ is $\tan x$.
$ \Rightarrow \sec x - \tan x + x + C$
Where $C$ is the arbitrary constant called the constant of integration.
Thus, the integration of $\dfrac{{\tan x}}{{\left( {\sec x + \tan x} \right)}}dx$ is, $\sec x - \tan x + x + C$ where $C$ is the integration constant.
So, the correct answer is “$\sec x - \tan x + x + C$”.

Note: We have indefinite integral that is why we added integration constant $C$. If we have a definite integral we do not add integration constant. There are different methods to solve indefinite integral questions. Here, in the given question we applied $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]} dx = \int {f\left( x \right)} dx \pm \int {g\left( x \right)} $ property. Suppose we have given any constant $k$ with the integral then we will apply another property: \[\int {\left[ {kf\left( x \right) \pm kg\left( x \right)} \right]} dx = k\int {f\left( x \right)} \pm k\int {g\left( x \right)} \]. The only difference between these two properties is, we kept the constant $k$ outside of integral.