Question

Integrate the following: \[\int{{{\cos }^{4}}2xdx}\].

Answer 1

Hint: For the above type of question we can see that it is difficult to integrate the function directly. So, we will use the trigonometric formulas which change the given term into such forms whose standard integration is known to us. The formula that we will use to change the given term into an integral form is as follows: -
\[{{\cos }^{2}}A=\dfrac{1+\cos 2A}{2}\]

Complete step-by-step answer:
We have been given an integral, \[\int{{{\cos }^{4}}2xdx}\] to evaluate. We can write the function as a square as shown below: -
\[\Rightarrow \int{{{\left( {{\cos }^{2}}2x \right)}^{2}}dx}\]
On using the formula of \[{{\cos }^{2}}2x\] in the above function we get,
\[{{\int{\left( \dfrac{1+\cos 4x}{2} \right)}}^{2}}dx\]
Now, we can take the square of the terms in the denominator as well as the numerator. We know that, \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]. So, we can expand the numerator using this. Then, we will get,
\[\int{\dfrac{1+{{\cos }^{2}}4x+2\cos 4x}{4}dx}\]
Now, we can take the constant terms outside the integral. Again, on substituting the value of \[{{\cos }^{2}}4x\] we get,
\[\Rightarrow \dfrac{1}{4}\int{\left( 1+\dfrac{1+\cos 8x}{2}+2\cos 4x \right)dx}\]
Now, on separating the each term of integral we get,
\[\Rightarrow \dfrac{1}{4}\left[ \int{1.dx}+\dfrac{1}{2}\int{\left( 1+\cos 8x \right)dx}+2\int{\cos 4x}.dx \right]\]
Applying the standard integral values, we get,
\[\Rightarrow \dfrac{1}{4}\left[ x+\dfrac{x+\dfrac{\sin 8x}{8}}{2}+2.\dfrac{\sin 4x}{4} \right]+C\]
Simplifying it further, we get,
\[\Rightarrow \dfrac{1}{4}\left[ x+\dfrac{8x+\sin 8x}{16}+\dfrac{\sin 4x}{2} \right]+C\]
Here, ‘C’ is any arbitrary constant.
Therefore, the integration of the given question is equal to \[\dfrac{1}{4}\left[ x+\dfrac{8x+\sin 8x}{16}+\dfrac{\sin 4x}{2} \right]+C\].

Note: Adding ‘C’ the arbitrary constant to the indefinite integral is must so be careful while solving the indefinite integral. Also, remember the trigonometry formulae as well as the trigonometry identities as it will help us in these kinds of questions where the trigonometric terms are not in the integral form.