Evaluate the following integral: \[\int{\dfrac{\cos 2x-\cos 2a}{\cos x-\cos a}dx}\] .

Hint: We will first apply trigonometric properties of $\cos \theta $ given by $\cos 2\theta =2{{\cos }^{2}}\theta -1$ on the given integral which will help us to simplify the given integral and after that we will apply basic mathematics property like ${{a}^{2}}-{{b}^{2}}=\left( a+b \right).\left( a-b \right)$ , this will help to further simplify and then we can cut out the common terms from numerator and denominator which will leave us with basic integral of $\cos \theta $.

Complete step-by-step solution:
We will first apply the property of $\cos \theta $ , Now we know that:
$\cos 2\theta =2{{\cos }^{2}}\theta -1$
We will apply the above property on the numerator part of the given integral.
Now,
$\cos 2x=2{{\cos }^{2}}x-1$ and $\cos 2a=2{{\cos }^{2}}a-1$
Substituting these values in the given integral: \[\int{\dfrac{\cos 2x-\cos 2a}{\cos x-\cos a}dx}\]
\[\int{\dfrac{\left( 2{{\cos }^{2}}x-1 \right)-\left( 2{{\cos }^{2}}a-1 \right)}{\cos x-\cos a}dx}=\int{\dfrac{2{{\cos }^{2}}x-1-2{{\cos }^{2}}a+1}{\cos x-\cos a}dx}=\int{\dfrac{2{{\cos }^{2}}x-2{{\cos }^{2}}a}{\cos x-\cos a}dx}\]
Taking out the constant: $\int{a}\cdot f\left( x \right)dx=a\cdot \int{f}\left( x \right)dx$
We will then have: \[\int{\dfrac{2{{\cos }^{2}}x-2{{\cos }^{2}}a}{\cos x-\cos a}dx}=2\int{\dfrac{{{\cos }^{2}}x-{{\cos }^{2}}a}{\cos x-\cos a}dx}\text{ }..............\text{ Equation 1}\text{. }\]
Now we know that: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right).\left( a-b \right)$
Applying it in the equation 1: \[2\int{\dfrac{{{\cos }^{2}}x-{{\cos }^{2}}a}{\cos x-\cos a}dx}=2\int{\dfrac{\left( \cos x-\cos a \right).\left( \cos x+\cos a \right)}{\cos x-\cos a}dx}\]
After cancelling out the terms, we will be left with: \[2\int{\left( \cos x+\cos a \right)dx}\]
Now we know the basic property of integration when it is given in the following form: \[\int{\left( f\left( x \right)+g\left( x \right) \right)}dx=\int{f\left( x \right)dx+}\int{g\left( x \right)dx}\]
We get:\[2\int{\left( \cos x+\cos a \right)dx}=2\left[ \int{\cos xdx+\int{\cos adx}} \right]\]
We will now apply the following two formulas for integration:
$\begin{align}
  & \Rightarrow \int{\cos \theta d\theta =\sin \theta } \\
 & \Rightarrow \int{dx=x} \\
\end{align}$
After applying the above two formulas in our question:
\[2\left[ \int{\cos xdx+\int{\cos adx}} \right]=2\left[ \sin x+x\cos a \right]+C\]
Therefore the result will be: \[\int{\dfrac{\cos 2x-\cos 2a}{\cos x-\cos a}dx}=2\left[ \sin x+x\cos a \right]+C\]

Note: Please note that $\cos a$ will not be integrated as we are finding out the integral with respect to the variable x, hence cos a will remain constant and the integral will give x as an output in the latter part of the integral. Do not get confused between the following integrations $\int{\cos \theta d\theta =\sin \theta }$ and $\int{\sin \theta d\theta =-\cos \theta }$ ; students might make mistakes with the negative sign in the output.