Question

Find the value of
$\int{\cos (x)\cos (2x)\cos (3x)dx}$

Answer 1

Hint: In this question, a multiplication of cosine terms is to be integrated. As the integration of terms involving just one cosine can be done using standard formulae, we should try to convert the integrand into a sum of cosine terms to obtain the required answer.

Complete step-by-step answer:
Here, the integrand (which is to be integrated) is in the form of a product of cosine terms. Therefore, we should try to convert it into a sum of terms involving a single cosine term. We can use the formula for the cosine of sum and difference of two terms which states that
$\begin{align}
  & \cos (a+b)=\cos (a)\cos (b)-\sin (a)\sin (b) \\
 & \cos (a-b)=\cos (a)\cos (b)+\sin (a)\sin (b) \\
\end{align}$
Adding the above equations, we obtain
$\begin{align}
  & \cos (a+b)+\cos (a-b)=\cos (a)\cos (b)-\sin (a)\sin (b)+\left( \cos (a)\cos (b)+\sin (a)\sin (b) \right) \\
 & =2\cos (a)\cos (b) \\
 & \Rightarrow \cos (a)\cos (b)=\dfrac{\cos (a+b)+\cos (a-b)}{2}............(1.1) \\
\end{align}$
Taking a=2x and b=x in equation (1.1), we obtain
$\cos (x)\cos (2x)=\cos (2x)\cos (x)=\dfrac{\cos (2x+x)+\cos (2x-x)}{2}=\dfrac{\cos (3x)+\cos (x)}{2}.............(1.2)$
Using equation (1.2), we can write the integrand given in the question as
$\begin{align}
  & \int{\cos (x)\cos (2x)\cos (3x)dx}=\int{\left( \dfrac{\cos (3x)+\cos (x)}{2} \right)\cos (3x)dx} \\
 & =\dfrac{1}{2}\int{\left( {{\cos }^{2}}(3x)+\cos (x)\cos (3x) \right)dx}................(1.3) \\
\end{align}$
Now, again using equation (1.1) by taking a=3x and b=x, we obtain
$\cos (x)\cos (3x)=\cos (3x)\cos (x)=\dfrac{\cos (3x+x)+\cos (3x-x)}{2}=\dfrac{\cos (4x)+\cos (2x)}{2}.............(1.4)$
Also, from trigonometric formulas,
${{\cos }^{2}}(x)=\dfrac{1+\cos (2x)}{2}$. Therefore, ${{\cos }^{2}}(3x)=\dfrac{1+\cos (6x)}{2}...................(1.5)$
Using the value of equation (1.4) and (1.5) in equation (1.3), we get
$\begin{align}
  & \int{\cos (x)\cos (2x)\cos (3x)dx}=\dfrac{1}{2}\int{\left( {{\cos }^{2}}(3x)+\cos (x)\cos (3x) \right)dx} \\
 & =\dfrac{1}{2}\int{\left( \dfrac{1+\cos (6x)}{2}+\dfrac{\cos (4x)+\cos (2x)}{2} \right)dx=\dfrac{1}{4}\int{dx+\dfrac{1}{4}\int{\cos (6x)dx}}+\dfrac{1}{4}\int{\cos (4x)dx+\dfrac{1}{4}\int{\cos (2x)}}}................(1.5) \\
\end{align}$
Now, from the standard integration formulae, we know that
$\int{\cos (nx)dx}=\dfrac{\sin (nx)}{n}+C....................(1.6)$
Where C is an arbitrary constant.
Using this in equation (1.5), we obtain
\[\begin{align}
  & \int{\cos (x)\cos (2x)\cos (3x)dx}=\dfrac{1}{4}\int{dx+\dfrac{1}{4}\int{\cos (6x)dx}}+\dfrac{1}{4}\int{\cos (4x)dx+\dfrac{1}{4}\int{\cos (2x)}} \\
 & =\dfrac{x}{4}+\dfrac{\sin (6x)}{4\times 6}+\dfrac{\sin (4x)}{4\times 4}+\dfrac{\sin (2x)}{4\times 2}+C=\dfrac{x}{4}+\dfrac{\sin (6x)}{24}+\dfrac{\sin (4x)}{16}+\dfrac{\sin (2x)}{8}+C \\
\end{align}\]
Which is the required answer to the given question.

Note: Note that we should be careful that the sign of the term after integrating cos(nx) should be positve (+sin(nx)), however the derivative of cos(x) yields –sin(x). Thus, the last three terms in the final answer should also have a positive sign. Also, as C is an arbitrary constant, C in the final answer is the sum of the constants obtained from all the integrations carried out in the step before it.