Question

Find the antiderivative of \[{(\cos x)^2}\]?

Answer 1

Hint:Finding integral in just opposite of differentiation so for finding an integral you have to go reverse as in differentiation, it's quite easy when you get it. In this question you have to buy parts and you will have to use little bit of trigonometric identities as per requirement in the solution.

Formulae Used:
 \[\cos 2x = 2{\cos ^2}x - 1\]

Complete step by step solution:
The given question is \[{(\cos x)^2}\]

After applying integration sign we get:
\[
\Rightarrow \int {{{(\cos x)}^2}} dx \\
\Rightarrow \int {{{\cos }^2}xdx} \\
\]
Here we are using the formulae of “cos(2x)” and integrating the term because direct integration of the given term is not available, hence we need to convert it into the given form, on solving we get:
\[
\Rightarrow \cos 2x = 2{\cos ^2}x - 1 \\
\Rightarrow 2{\cos ^2}x = \cos 2x + 1 \\
\Rightarrow {\cos ^2}x = \dfrac{1}{2}\left( {\cos 2x + 1} \right) \\
\]
Now we are using this conversion in our equation to solve the integral, and accordingly we will find the integral of the given term,

Here we get,
\[
\Rightarrow \int {\dfrac{1}{2}\left( {\cos 2x + 1} \right)dx = \dfrac{1}{2}\left( {\dfrac{1}{2}\sin 2x +
x} \right)} \left( {\int {\cos 2x = \dfrac{1}{2}\sin 2x} } \right) \\
\Rightarrow \dfrac{1}{2}\left( {\dfrac{1}{2}\sin 2x + x} \right) = \dfrac{1}{4}\sin 2x + \dfrac{x}{2}
\\
\]
This is our required solution for the given equation.

Additional Information: You can check the answer by solving from bottom to top; just you have to differentiate the last equation until you get the fist equation. This method can take some time and several steps more than you take for integration but definitely you can get and verify.

Note: This is the easiest way to solve this question, but if you find any difficulty then you can try another method that is, use expansion of \[{\cos ^n}x\]and after expansion you can integrate each term independently, then you can merge it and obtain your result. This method is appropriate for this kind of question.