Question

How do you find integral for the following expression\[\int {\dfrac{{\sec x\tan x}}{{\sec x - 1}}dx} \]?

Answer 1

Hint: The above question deals with the concept of integration. Integration, at its basic, is the inverse process of differentiation. Unlike differentiation, in integration, you are given a derivative and you have to find its primitive that is the original function. In order to solve the given question, we will first change the values of trigonometric functions using the reciprocal and quotient trigonometric identities. After which we will use the substitution method to solve it further.

Complete step-by-step solution:
Given that,
 \[ \Rightarrow \int {\dfrac{{\sec x\tan x}}{{\sec x - 1}}dx} \]
According to the reciprocal and quotient trigonometric identities, we know that,
\[ \Rightarrow \sec x = \dfrac{1}{{\cos x}}\] and \[ \tan x = \dfrac{{\sin x}}{{\cos x}}\]
Therefore, we put these values above in our given expression,
We get,
\[ \Rightarrow \int {\dfrac{1}{{\cos x}} \times \dfrac{{\sin x}}{{\cos x}} \times \dfrac{1}{{\dfrac{1}{{\cos x}} - 1}}dx} \]
Therefore,
\[ \Rightarrow \int {\dfrac{{\sin x}}{{\cos x - {{\cos }^2}x}}dx} - - - - - (1)\]
Now applying the substitution method,
Let \[u = \cos x - - - - - \left( 2 \right)\]
So,
\[ \Rightarrow du = - \sin x\]
Now, substituting this value in equation (1)
We get,
\[ \Rightarrow \int {\dfrac{1}{{u - {u^2}}}( - du)} \]
Thus,
\[ \Rightarrow - \int {\dfrac{1}{{u(1 - u)}}du} \]
Splitting the expression and taking integration separately,
\[ \Rightarrow - \int {\dfrac{1}{u}du} + \int {\dfrac{1}{{1 - u}}du} \]
Thus, we get
\[ \Rightarrow - \left| {\log u} \right| + \left| {\log (1 - u)} \right|\]
Now, putting the value of u from equation (2)
\[ \Rightarrow - \left| {\log \cos (x)} \right| + \left| {\log (1 - \cos x)} \right|\]
Therefore,
\[ \Rightarrow 2\log \left\{ {\sin \left( {\dfrac{x}{2}} \right)} \right\} - \log \left( {\cos x} \right) + C\]
Which is our required integral.

Additional information: Another way of doing the same question is as follows:
We have,
\[ \Rightarrow \int {\dfrac{{\sec x\tan x}}{{\sec x - 1}}dx} \]
We know that the derivation of \[\sec x\]is \[\sec x\tan x\]
Therefore our expression becomes,
\[ \Rightarrow \int {\dfrac{{\sec x\tan x}}{{\sec x - 1}}dx} = \int {d\dfrac{{\sec }}{{\sec x - 1}}} \]
Here let \[u = \sec x\]
Thus we get,
\[ \Rightarrow \int {\dfrac{1}{{u - 1}}du} \]
\[ \Rightarrow \left| {\log (1 - u)} \right| + C\]
Putting the value of u,
\[ \Rightarrow \log \left| {\sec x - 1} \right| + C\]

Note: In general, there can be more than one way of solving integration. The method that we used in the above question is the substitution method. Other than this, there are two more methods of solving integration, namely, integration by parts and integration by partial fraction method. One might have to use a combination of methods to solve integrations.