Question

Find the antiderivative of ${\cos ^2}\left( x \right)$.

Answer 1

Hint:
To find the antiderivative of any function the opposite of differentiation (finding the derivative) needs to be done, i.e. the given function needs to be integrated.

Complete step by step solution:
To find the antiderivative of the function ${\cos ^2}\left( x \right)$ we need to integrate it.
To integrate the function, at first, it should be converted to a simpler form that can be directly integrated. To do so a multiple angle formula can be used.
We know
$\cos \left( {2x} \right) = {\cos ^2}\left( x \right) - {\sin ^2}\left( x \right)$ ……………..(1)
Also we know the trigonometric identity
${\cos ^2}\left( x \right) + {\sin ^2}\left( x \right) = 1$
$ \Rightarrow {\sin ^2}\left( x \right) = 1 - {\cos ^2}\left( x \right)$ ………………….(2)
Substituting ${\sin ^2}x$from (2) in (1) :
$ \Rightarrow \cos \left( {2x} \right) = {\cos ^2}\left( x \right) - \left( {1 - {{\cos }^2}\left( x \right)} \right)$
Opening the bracket:
$ \Rightarrow \cos \left( {2x} \right) = {\cos ^2}\left( x \right) - 1 + {\cos ^2}\left( x \right)$
$ \Rightarrow \cos \left( {2x} \right) = 2{\cos ^2}\left( x \right) - 1$
Changing sides:
$ \Rightarrow 2{\cos ^2}\left( x \right) = \cos \left( {2x} \right) + 1$
$ \Rightarrow {\cos ^2}\left( x \right) = \dfrac{{\cos \left( {2x} \right) + 1}}{2}$
Hence, integrating both sides of the above integration
$\int {{{\cos }^2}} \left( x \right)dx = \int {\dfrac{{\cos \left( {2x} \right) + 1}}{2}} $
Splitting the right hand side term:
\[ \Rightarrow \int {{{\cos }^2}\left( x \right)} dx = \dfrac{1}{2}\int {\cos \left( {2x} \right)} dx + \dfrac{1}{2}\int {1dx} \]
Since, \[\int {\cos \left( {nx} \right)} dx = \dfrac{{\sin \left( {nx} \right)}}{n} + C\], where \[n \in R\] and \[\int {1dx = x + C} \]
\[ \Rightarrow \int {{{\cos }^2}\left( x \right)} dx = \dfrac{1}{2}\dfrac{{\sin \left( {2x} \right)}}{2} + \dfrac{1}{2}x + C\]
Where \[C\] is an arbitrary constant of integration.
\[ \Rightarrow \int {{{\cos }^2}\left( x \right)} dx = \dfrac{1}{4}\sin \left( {2x} \right) + \dfrac{1}{2}x + C\]

Hence the antiderivative of \[{\cos ^2}x\] is equal to \[\dfrac{1}{4}\sin \left( {2x} \right) + \dfrac{1}{2}x + C\].

Additional information:
If in case there is any function that cannot be converted to any simpler form then another approach is to try integration by parts.

Note:
For functions like this where any direct formula of integration cannot be applied, the first approach should be to convert it into a simpler format using trigonometric formulas and identities such that the terms can be directly integrated.