Question

How do you find the integral of $\int {{{(\cos x)}^4}} $.

Answer 1

Hint: In this question we will make appropriate substitutions for the given terms to simplify the expression into simplified integrals.

Formula used: ${\cos ^2}x = \dfrac{{1 + \cos (2x)}}{2}$

Complete step-by-step solution:
We have the expression as: $\int {{{(\cos x)}^4}} dx$
It can be written as a product of two terms as:
$ \Rightarrow \int {{{(\cos x)}^2} \times } {(\cos x)^2}dx$
This can be written as:
$ \Rightarrow \int {{{\cos }^2}x \times } {\cos ^2}xdx$
Now we know that:
$ \Rightarrow {\cos ^2}x = \dfrac{{1 + \cos (2x)}}{2}$ therefore, on substituting it in the above expression we get:
\[ \Rightarrow \int {\dfrac{{1 + \cos (2x)}}{2} \times \dfrac{{1 + \cos (2x)}}{2}} dx\]
Now on taking the common multiple value out of the integral, we get:
\[ \Rightarrow \dfrac{1}{4}\int {(1 + \cos (2x))(1 + \cos (2x)} )dx\]
On multiplying the terms, we get:
\[ \Rightarrow \dfrac{1}{4}\int {1 + 2\cos (2x) + {{\cos }^2}(2x)} dx\]
Now on using the formula again in the above term, we get:
\[ \Rightarrow \dfrac{1}{4}\int {1 + 2\cos (2x) + \dfrac{{1 + \cos (4x)}}{2}} dx\]
Now on taking the lowest common multiple on all the terms, we get:
\[ \Rightarrow \dfrac{1}{4}\int {\dfrac{{2 + 4\cos (2x) + 1 + \cos (4x)}}{2}} dx\]
Now on taking the common multiple out of the integral, we get:
\[ \Rightarrow \dfrac{1}{8}\int {2 + 4\cos (2x) + 1 + \cos (4x)} dx\]
Now on simplifying the expression, we get:
\[ \Rightarrow \dfrac{1}{8}\int {3 + 4\cos (2x) + \cos (4x)} dx\]
Now we will distribute the integral and solve the integrals using the formulae of integration.
On distributing the integrals, we get:
\[ \Rightarrow \dfrac{1}{8}\int {3 + \int {4\cos (2x)} + \int {\cos (4x)} } dx\]
On removing the multiple out of the integral, we get:
\[ \Rightarrow \dfrac{1}{8}\int {3 + 4\int {\cos (2x)} + \int {\cos (4x)} } dx\]
Now we know that $\int k = kx + c$ and $\int {\cos (ax) = \dfrac{1}{a}\sin (ax)} $
Therefore, on using these formulae, we get:
\[ \Rightarrow \dfrac{1}{8}\left[ {3x + 4\dfrac{{\sin (2x)}}{2} + \dfrac{{\sin (4x)}}{4}} \right] + c\]
On taking the lowest common multiple, we get:
\[ \Rightarrow \dfrac{1}{8}\left[ {\dfrac{{12x}}{4} + \dfrac{{8\sin (2x)}}{4} + \dfrac{{\sin (4x)}}{4}} \right] + c\]
On taking the common multiple out of the fraction, we get:
\[ \Rightarrow \dfrac{1}{{32}}\left[ {12x + 8\sin (2x) + \sin (4x)} \right] + c\]

\[\dfrac{1}{{32}}\left[ {12x + 8\sin (2x) + \sin (4x)} \right] + c\] is the required solution.

Note: It is to be remembered that integration and derivatives is the inverse of each other. If the derivative of $a$ is $b$ then the integration of $b$ is $a$.
Unlike derivatives, integration does not have a chain rule to find out the integration.
In some questions there might be multiplication of two terms, in those types of questions the integration by parts formula should be used which is:
$\int {uvdx} = u\int v dx - \int {\left( {\dfrac{{du}}{{dx}}\int {vdx} } \right)} $Where $u$ and $v$ are the terms in multiplication.