Question

How do you find the integral of $\int {{{\left( {\cos x} \right)}^4}dx} $?

Answer 1

Hint: First square the cosine and apply the formula ${\cos ^2}x = \dfrac{{1 + \cos x}}{2}$. After that square the expression. Then again apply the above formula on ${\cos ^2}2x$. After that simplify the terms. Then distribute the integral on each term and do a simple integration to get the desired result.

Complete step-by-step answer:
We are asked to integrate the given function $\int {{{\left( {\cos x} \right)}^4}dx} $.
The terms can be rewritten as,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \int {{{\left( {{{\cos }^2}x} \right)}^2}dx} $
We know that,
${\cos ^2}x = \dfrac{{1 + \cos x}}{2}$
Substitute the above formula in the integration,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \int {{{\left( {\dfrac{{1 + \cos 2x}}{2}} \right)}^2}dx} $
Now square the terms by the formula ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \int {\dfrac{{1 + 2\cos 2x + {{\cos }^2}2x}}{4}dx} $
Take out $\dfrac{1}{4}$ outside of the integration,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{4}\int {\left( {1 + 2\cos 2x + {{\cos }^2}2x} \right)dx} $
Again, apply the formula on ${\cos ^2}2x$,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{4}\int {\left( {1 + 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right)dx} $
Take LCM on the right side,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{4}\int {\left( {\dfrac{{2 + 4\cos 2x + 1 + \cos 4x}}{2}} \right)dx} $
Add the like terms and take out $\dfrac{1}{2}$ outside of the integration,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{8}\int {\left( {3 + 4\cos 2x + \cos 4x} \right)dx} $
Now, distribute the integral,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{8}\left( {\int {3dx} + \int {4\cos 2xdx} + \int {\cos 4xdx} } \right)$
Take out the constant term,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{8}\left( {3\int {1dx} + 4\int {\cos 2xdx} + \int {\cos 4xdx} } \right)$
Now, integrate the terms,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{8}\left( {3x + 4 \times \dfrac{{\sin 2x}}{2} + \dfrac{{\sin 4x}}{4}} \right)$
Now take LCM inside the bracket,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{1}{8}\left( {\dfrac{{12x + 8\sin 2x + \sin 4x}}{4}} \right)$
Open bracket and multiply the denominator,
$ \Rightarrow \int {{{\left( {\cos x} \right)}^4}dx} = \dfrac{{12x + 8\sin 2x + \sin 4x}}{{32}}$

Hence, the integral of $\int {{{\left( {\cos x} \right)}^4}dx} $ is $\dfrac{{12x + 8\sin 2x + \sin 4x}}{{32}}$.

Note:
Students should keep in mind the formula of finding integration of the trigonometric function. Students should note that we always need to simplify our function as there exists no formula for finding integration of the two dividing functions.
We split the function inside the integral only because the operation between them is added. If the operation is multiplication then we cannot split them; we need to apply integration by parts.
According to integration definition, it is a process of finding functions whose derivative is given is named anti-differentiation or integration. Integration is a process of adding slices to find the whole. It can be used to find areas, volumes, and central points.