What is $ \int{{{\cos }^{2}}\left( 6x \right)dx} $ ?

Hint: We first use the trigonometric form of multiple angles where $ {{\cos }^{2}}z=\dfrac{1+\cos 2z}{2} $ to simplify the function form of $ {{\cos }^{2}}\left( 6x \right) $ . Then we break the integration in two parts and use the integral form of $ \int{\cos mzdz}=\dfrac{\sin mz}{m} $ and $ \int{dz}=z $ . We use an integral constant to find the final solution.

Complete step by step solution:
We need to convert the trigonometric form of $ {{\cos }^{2}}\left( 6x \right) $ using the form of $ {{\cos }^{2}}z=\dfrac{1+\cos 2z}{2} $ .
We assume $ z=6x $ and get $ {{\cos }^{2}}\left( 6x \right)=\dfrac{1+\cos \left( 12x \right)}{2} $ .
Now we integrate the function separately and get
$ \int{{{\cos }^{2}}\left( 6x \right)dx}=\int{\left\{ \dfrac{1+\cos \left( 12x \right)}{2} \right\}dx} $ .
The function gets separated and gives
$ \int{\left\{ \dfrac{1+\cos \left( 12x \right)}{2} \right\}dx}=\dfrac{1}{2}\int{dx}+\dfrac{1}{2}\int{\cos \left( 12x \right)dx} $ .
We know that $ \int{\cos mzdz}=\dfrac{\sin mz}{m} $ and $ \int{dz}=z $ .
Therefore, the integral form is
$ \int{\left\{ \dfrac{1+\cos \left( 12x \right)}{2} \right\}dx}=\dfrac{x}{2}+\dfrac{\sin \left( 12x \right)}{24}+c $ .
Here $ c $ is the integral constant.
Therefore, the solution of $ \int{{{\cos }^{2}}\left( 6x \right)dx} $ is $ \dfrac{x}{2}+\dfrac{\sin \left( 12x \right)}{24}+c $ .
So, the correct answer is “$ \dfrac{x}{2}+\dfrac{\sin \left( 12x \right)}{24}+c $ ”.

Note: We can also solve the integration using the base change for the variable $ z=\cos x $ . In that case the sum gets complicated but the final solution would be the same. It is better to watch out for the transformations in the trigonometric forms in the function and take that as the change in the variable.