Question

How do you find the integral of $\cos \left( mx \right).\cos \left( nx \right)$ ?

Answer 1

Hint: We have been given the product of the same trigonometric function, cosine function but bearing two different angles as a function of the same x-variable. In order to solve such problems, we must have prior knowledge of basic trigonometric identities. Thus, we shall apply a trigonometric identity to a given function to simplify it further and we shall integrate it accordingly.

Complete step-by-step solution:
We are given the function, $\cos \left( mx \right).\cos \left( nx \right)$.
We know that one of the basic trigonometric properties convert the product of the cosine functions into the sum of cosine functions. It is expressed as:
$\cos A.\cos B=\dfrac{1}{2}\left[ \cos \left( A-B \right)+\cos \left( A+B \right) \right]$
Here, we have angle $A=mx$ and angle $B=nx$. Applying this identity on given function, we get
$\cos \left( mx \right).\cos \left( nx \right)=\dfrac{1}{2}\left[ \cos \left( mx-nx \right)+\cos \left( mx+nx \right) \right]$
Taking x common in both the terms in the right-hand side, we get
$\Rightarrow \cos \left( mx \right).\cos \left( nx \right)=\dfrac{1}{2}\left[ \cos \left( m-n \right)x+\cos \left( m+n \right)x \right]$
We shall now integrate this modified expression to find the final solution.
$\Rightarrow \int{\cos \left( mx \right).\cos \left( nx \right)}=\int{\dfrac{1}{2}\left[ \cos \left( m-n \right)x+\cos \left( m+n \right)x \right]}$
Since $\dfrac{1}{2}$ is constant term, thus taking it out of integration function, we get
\[\Rightarrow \int{\cos \left( mx \right).\cos \left( nx \right)}=\dfrac{1}{2}\int{\cos \left( m-n \right)x+\cos \left( m+n \right)x}\]
By the basic properties of integration, we know that $\int{\cos ax.dx=\dfrac{\sin ax}{a}+C}$. Applying this property, we get
\[\Rightarrow \int{\cos \left( mx \right).\cos \left( nx \right)}=\dfrac{1}{2}\left[ \dfrac{\sin \left( m-n \right)x}{m-n}+C+\dfrac{\sin \left( m+n \right)x}{m+n}+C \right]\]
Taking the constants of integration $\left( C \right)$ common, we get
\[\Rightarrow \int{\cos \left( mx \right).\cos \left( nx \right)}=\dfrac{1}{2}\left[ \dfrac{\sin \left( m-n \right)x}{m-n}+\dfrac{\sin \left( m+n \right)x}{m+n} \right]+C\]
 Now, we shall multiply the $\dfrac{1}{2}$ outside the brackets with both the terms individually,
 \[\Rightarrow \int{\cos \left( mx \right).\cos \left( nx \right)}=\dfrac{\sin \left( m-n \right)x}{2\left( m-n \right)}+\dfrac{\sin \left( m+n \right)x}{2\left( m+n \right)}+C\]
Therefore, the integral of $\cos \left( mx \right).\cos \left( nx \right)$ is \[\dfrac{\sin \left( m-n \right)x}{2\left( m-n \right)}+\dfrac{\sin \left( m+n \right)x}{2\left( m+n \right)}+C\].

Note: The constant of integration represents indefinite integration of a function. This is when the limits of integration are not specified. Another method of solving this problem is by applying integration by parts. This is done by using the ILATE rule which is used when the product of two separately integrable functions is given. According to the preference order of the ILATE rule, we take one of the functions as the first function which must be easily integrable and the second function must be easily differentiable.