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Evaluate \[\int {\dfrac{{\left( {1 - \cos x} \right)dx}}{{\cos x\left( {1 + \cos x} \right)}}} \]
A. \[\log \left( {\sec x + \tan x} \right) - 2\tan \dfrac{x}{2} + c\]
B. \[\log \left( {\sec x + \tan x} \right) + 2\tan \dfrac{x}{2} + c\]
C. \[\log \left( {\sec x - \tan x} \right) - 2\tan \dfrac{x}{2} + c\]
D. \[\log \left( {\sec x - \tan x} \right) + 2\tan \dfrac{x}{2} + c\]

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Last updated date: 18th Jun 2024
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Answer
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Hint: Here, we will use the trigonometric identity and inverse trigonometric ratio to simplify the integrand. We will substitute the variables for the simplified integrand. Then by using the concept of integration, we will integrate the function. Integration is defined as the summation of all the discrete data.

Formula Used:
We will use the following formula:
1. Trigonometric Identity: \[\cos x = 2{\cos ^2}\dfrac{x}{2} - 1\] and\[\dfrac{1}{{\cos x}} = \sec x\]
2. Derivative Formula: \[\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\],\[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\] and \[\dfrac{d}{{dx}}\left( x \right) = 1\]
3. Integral Formula: \[\int {\dfrac{1}{u}} du = \ln \left( u \right)\] and\[\int {{{\sec }^2}vdv = \tan v} \]

Complete Step by Step Solution:
We are given that the integral function \[\int {\dfrac{{\left( {1 - \cos x} \right)dx}}{{\cos x\left( {1 + \cos x} \right)}}} \]
Let the given integral function be \[I\].
\[I = \int {\dfrac{{\left( {1 - \cos x} \right)dx}}{{\cos x\left( {1 + \cos x} \right)}}} \]
Now, we will rewrite the integrand in the numerator, we get
\[ \Rightarrow I = \int {\dfrac{{\left( {1 + \cos x - 2\cos x} \right)dx}}{{\cos x\left( {1 + \cos x} \right)}}} \]
Now, by segregating the Integrand, we get
\[ \Rightarrow I = \int {\dfrac{{1 + \cos x}}{{\cos x\left( {1 + \cos x} \right)}} - \dfrac{{2\cos x}}{{\cos x\left( {1 + \cos x} \right)}}dx} \]
Now, by cancelling the common terms in the numerator and the denominator, we get
\[ \Rightarrow I = \int {\dfrac{1}{{\cos x}} - \dfrac{2}{{\left( {1 + \cos x} \right)}}dx} \]
By using the Trigonometric Identity \[\cos x = 2{\cos ^2}\dfrac{x}{2} - 1\] and \[\dfrac{1}{{\cos x}} = \sec x\], we get
\[ \Rightarrow I = \int {\sec x - \dfrac{2}{{2{{\cos }^2}\dfrac{x}{2}}}dx} \]
Now, cancelling out the common terms, we get
\[ \Rightarrow I = \int {\sec x - \dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}}dx} \]
Using the Trigonometric Identity \[\dfrac{1}{{\cos x}} = \sec x\], we get
\[ \Rightarrow I = \int {\sec x - {{\sec }^2}\dfrac{x}{2}dx} \]
Now, by segregating the Integrand, we get
\[ \Rightarrow I = \int {\sec xdx - \int {{{\sec }^2}\dfrac{x}{2}dx} } \]
Now, by multiplying \[\left( {\sec x + \tan x} \right)\] to the numerator and the denominator of the first integral function, we get
\[ \Rightarrow I = \int {\sec x \times \dfrac{{\left( {\sec x + \tan x} \right)}}{{\left( {\sec x + \tan x} \right)}}dx - \int {{{\sec }^2}\dfrac{x}{2}dx} } \]
Now, by multiplying the terms, we get
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}x + \sec x\tan x}}{{\left( {\sec x + \tan x} \right)}}dx - \int {{{\sec }^2}\dfrac{x}{2}dx} } \]……………………………….\[\left( 1 \right)\]
Let us consider \[u = \sec x + \tan x\]
Now, we will differentiate the variable \[u\] by using the derivative formula \[\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x\] and \[\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x\]. Therefore, we get
\[ \Rightarrow du = \left( {\sec x\tan x + {{\sec }^2}x} \right)dx\] .
Let us consider \[v = \dfrac{x}{2} = \dfrac{1}{2}x\]
Now, we will differentiate the variable \[v\] by using the derivative formula \[\dfrac{d}{{dx}}\left( x \right) = 1\], so we get
\[\begin{array}{l} \Rightarrow dv = \dfrac{1}{2}dx\\ \Rightarrow dx = 2dv\end{array}\]
Substituting \[dx = 2dv\] and \[du = \left( {\sec x\tan x + {{\sec }^2}x} \right)dx\] in equation \[\left( 1 \right)\], we get
\[I = \int {\dfrac{1}{u}du - 2\int {{{\sec }^2}vdv} } \]
Now, by using Integral formula \[\int {\dfrac{1}{u}} du = \ln \left( u \right)\] and\[\int {{{\sec }^2}vdv = \tan v} \] , we will integrate the function.
\[ \Rightarrow I = \ln \left( u \right) - 2\tan v + c\]
By substituting the variables \[u\] and \[v\] , we get
\[ \Rightarrow I = \ln \left( {\sec x + \tan x} \right) - 2\tan \dfrac{x}{2} + c\]
\[ \Rightarrow I = \log \left( {\sec x + \tan x} \right) - 2\tan \dfrac{x}{2} + c\]
Therefore, the value of \[\int {\dfrac{{\left( {1 - \cos x} \right)dx}}{{\cos x\left( {1 + \cos x} \right)}}} \] is \[\log \left( {\sec x + \tan x} \right) - 2\tan \dfrac{x}{2} + c\].

Thus, option(A) is the correct answer.

Note:
We know that Integration is the process of adding the small parts to find the whole parts. We need to keep in mind that whenever we are using the method of substitution in integration, the variable to be integrated also changes according to the substitution. 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 are no limits in the integral. Whenever the integration is done with no limits, then an Arbitrary constant should be added.