Question

Find the integral of \[\int {\cos x\cos 2x\cos 3xdx} \]

Answer 1

Hint: According to the given question, firstly we will try to make trigonometric identity by multiplying and dividing the equation with 2 so that we can use the formula of \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\] and convert it into simpler form. Then, again on multiplying and dividing the equation with 2 we can use the above identity and to make it more simpler we can also use the identity of \[2{\cos ^2}x = 1 + \cos 2x\] and integrate the simplified result.
Formula used: Here we will use the trigonometric formulas of \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\] and \[2{\cos ^2}x = 1 + \cos 2x\] .

Complete step-by-step answer:
As it is given that we have to evaluate \[\int {\cos x\cos 2x\cos 3xdx} \]
Multiply and divide the equation with 2. So, we can apply the formula of \[2\cos A\cos B\] .
 \[ \Rightarrow \dfrac{1}{2}\int {\left( {2\cos x\cos 2x} \right)\cos 3xdx} \]
Now using the formula of \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\] .
As Here, \[A = \cos x\] and \[B = \cos 2x\] after substituting these values in the equation we get,
 \[ \Rightarrow \dfrac{1}{2}\int {\left( {\cos 3x + \cos x} \right)\cos 3xdx} \]
By opening brackets and then multiplying the above equation we get,
 \[ \Rightarrow \dfrac{1}{2}\int {{{\cos }^2}3x + \cos x\cos 3xdx} \]
Again multiplying and dividing the equation with 2 to make it the identity of \[2\cos A\cos B\] .
 \[ \Rightarrow \dfrac{1}{{2 \times 2}}\int {2\left( {{{\cos }^2}3x + \cos x\cos 3x} \right)dx} \]
Opening brackets and then multiplying with each other. So, we get
 \[ \Rightarrow \dfrac{1}{4}\int {2{{\cos }^2}3x + 2\cos x\cos 3xdx} \]
Again using the formula of \[2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)\] .
As Here, \[A = \cos x\] and \[B = \cos 3x\] after substituting these values in the equation we get, .
 \[ \Rightarrow \dfrac{1}{4}\int {2{{\cos }^2}3x + \cos 4x + \cos 2xdx} \]
Opening \[2{\cos ^2}3x\] by using the identity \[2{\cos ^2}x = 1 + \cos 2x\] here \[x = 3x\] then we get,
 \[ \Rightarrow \dfrac{1}{4}\int {1 + \cos 6x + \cos 4x + \cos 2xdx} \]
Now, we will integrate with the above equation with respect to x .
 \[ \Rightarrow \dfrac{1}{4}\left[ {x + \dfrac{{\sin 6x}}{6} + \dfrac{{\sin 4x}}{4} + \dfrac{{\sin 2x}}{2}} \right] + c\] which is our required result.
So, the correct answer is “ \[ \Rightarrow \dfrac{1}{4}\left[ {x + \dfrac{{\sin 6x}}{6} + \dfrac{{\sin 4x}}{4} + \dfrac{{\sin 2x}}{2}} \right] + c\] ”.

Note: To solve these types of questions, we must remember the trigonometric formulas to solve it in a simple way. Once it is simplified we can easily integrate it as we have the equation in the simplest trigonometric functions.