Question

Solve the following integration: $\int{\cos 2x\cos 4x\cos 6xdx}$.

Answer 1

Hint: First try to convert the term inside the integral into simple terms which you can solve easily. The terms which you know the integration. First, try to convert the product into a sum of different terms. If it is the sum you can integrate separately and then add them to get the result. This is the simplest possible way because you are breaking the whole large term into chunks of small terms. Here use the formulae of transformation to make it simpler. Then you get the sum of products of 2 terms. So, again apply the same formula to convert it into simpler.
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$

Complete step-by-step solution -
Given integral in the question which we need to solve:
$\int{\cos 2x\cos 4x\cos 6xdx}$
Assume the integral value to be variable I, we get:
$\Rightarrow I=\int{\cos 2x\cos 4x\cos 6xdx}$
Multiply and divide with 2 on right hand side of I, we get:
$\Rightarrow I=\dfrac{2}{2}\int{\cos 2x\cos 4x\cos 6x\text{ }dx}$
By sending2 inside the integral, we can write it in the form:
$\Rightarrow I=\dfrac{1}{2}\int{\left( 2\cos 2x\cos 4x \right)\cos 6x\text{ }dx}$
By basic knowledge of transformation, we know that the formula:
$2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$
By applying the above formula, we get the value of I as:
$\Rightarrow I=\dfrac{1}{2}\int{\left( \cos \left( 2x+4x \right)+\cos \left( 4x-2x \right) \right)\cos 6x\text{ }dx}$
By simplifying the above equation, we get the value of I as:
$\Rightarrow I=\dfrac{1}{2}\int{\cos 6x\cos 6x+\cos 2x\cos 6x\text{ }dx}$
Again, multiply and divide by 2 and send 2 inside the integral:
$\Rightarrow I=\dfrac{1}{4}\int{2\left( \cos 6x\cos 6x+\cos 2x\cos 6x \right)dx}$
By simplifying the above equation, we get value of I as:
$\Rightarrow I=\dfrac{1}{4}\int{\left( 2\cos 6x\cos 6x+2\cos 2x\cos 6x \right)dx}$
By applying the same transformation formula, we get value of I as:
$\Rightarrow I=\dfrac{1}{4}\int{\left( \cos \left( 6x+6x \right)+\cos \left( 6x-6x \right)+\cos \left( 6x+2x \right)+\cos \left( 6x-2x \right) \right)dx}$
By separating each term and adding them separately, we get:
$\Rightarrow I=\dfrac{1}{4}\int{\cos 12xdx}+\dfrac{1}{4}\int{\cos 0dx}+\dfrac{1}{4}\int{\cos 8xdx}+\dfrac{1}{4}\int{\cos 4xdx}$
Use the formulae of integration: $\int{\cos \left( ax+b \right)dx=\dfrac{\sin \left( ax+b \right)}{a}+c,\int{1dx=x+c}}$
By substituting them we get the value of I to be:
$\Rightarrow I=\dfrac{1}{4}\left[ \dfrac{\sin 12x}{12}+x+\dfrac{\sin 8x}{8}+\dfrac{\sin 4x}{4} \right]+c$
Therefore, I is the value of the given integral.

Note: Alternate method is write $\left( \cos 6x\cdot \cos 6x \right)$ as ${{\cos }^{2}}6x$ and use the formula $\cos 2\theta =2{{\cos }^{2}}\theta -1$ and solve. Anyways, you will get the same result. Be careful while taking $\cos 0$. Generally, students miss this term as it is constant but inside integral constants give us the x in the result, while will matter to our result.