Question

Integrate $\int {{{\cos }^4}xdx}$

Answer 1

We will write the integrand as a square of ${\cos ^2}x$. Then we will use an appropriate trigonometric identity to replace ${\cos ^2}x$ in order to make it easily integrable. Finally, we will simplify the expression and integrate each term using the formula to get the required answer.

Formula used:
We will use the following formulas:

1. $\cos 2x = 2{\cos ^2}x - 1$
2. $\int {kf(x)dx = k\int {f(x)dx} } + c$
3. $\int {\cos nxdx = \dfrac{{\sin nx}}{n} + c}$

Complete step by step solution:
We have to integrate $\int {{{\cos }^4}xdx}$. Let us denote this integral by $I$.
$I = \int {{{\cos }^4}xdx}$
Now, the integrand is not in an integrable form.
We will use the identity $\cos 2x = 2{\cos ^2}x - 1$ to bring it into an easily integrable form. From the identity we get
${\cos ^2}x = \dfrac{{1 + \cos 2x}}{2}$ ………$\left( 1 \right)$
We can write ${\cos ^4}x = {({\cos ^2}x)^2}$.
Let us substitute the equation $\left( 1 \right)$ in place of ${\cos ^2}x$. Thus, we get
${\cos ^4}x = {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)^2}$
We will now expand the terms on the RHS using the identity ${(a + b)^2} = {a^2} + 2ab + {b^2}$. So, the above equation becomes
$\Rightarrow {\cos ^4}x = \dfrac{1}{4}\left[ {{1^2} + 2 \times 1 \times \cos 2x + {{(\cos 2x)}^2}} \right]$
Simplifying the expression, we get
$\Rightarrow {\cos ^4}x = \dfrac{1}{4}\left[ {1 + 2\cos 2x + {{\cos }^2}2x} \right]$ ………$\left( 2 \right)$
We see that on the RHS, there is a term ${\cos ^2}2x$. Using equation $\left( 1 \right)$, we can write this as
${\cos ^2}2x = \dfrac{{1 + \cos 2.2x}}{2} = \dfrac{{1 + \cos 4x}}{2}$
Substituting this in equation $\left( 2 \right)$, we have
{\cos ^4}x = \dfrac{1}{4}\left\\{ {1 + 2\cos 2x + \dfrac{{1 + \cos 4x}}{2}} \right\\}
Now, we will take LCM on the RHS. So, we have
\Rightarrow {\cos ^4}x = \dfrac{1}{4}\left\\{ {\dfrac{{2 + 4\cos 2x + 1 + \cos 4x}}{2}} \right\\}
\Rightarrow {\cos ^4}x = \dfrac{1}{8}\left\\{ {3 + 4\cos 2x + \cos 4x} \right\\}
We see that each term of the above equation is in an integrable form. Now, we can easily integrate using integration formulae. Thus,
I = \int {{{\cos }^4}xdx} = \int {\dfrac{1}{8}\left\\{ {3 + 4\cos 2x + \cos 4x} \right\\}dx}
Using the formula $\int {kf(x)dx = k\int {f(x)dx} } + c$ to bring the constant $\dfrac{1}{8}$ outside the integral, we get,
I = \dfrac{1}{8}\int {\left\\{ {3 + 4\cos 2x + \cos 4x} \right\\}dx} ……….$\left( 3 \right)$
Now, we will integrate each term separately on the RHS.
Using the formula $\int {kdx = kx + } c$,$\int {\cos nxdx = \dfrac{{\sin nx}}{n} + c}$, and $\int {\cos 4xdx = \dfrac{{\sin 4x}}{4} + c}$, the equation $\left( 3 \right)$ becomes
$\Rightarrow I = \dfrac{1}{8}\left[ {3x + 4\dfrac{{\sin 2x}}{2} + \dfrac{{\sin 4x}}{4}} \right] + c$
$\Rightarrow I = \dfrac{1}{8}\left[ {3x + 2\sin 2x + \dfrac{{\sin 4x}}{4}} \right] + c$
We will simplify further by multiplying $\dfrac{1}{8}$ to each of the terms inside the brackets on the RHS. Therefore, we get

$\Rightarrow I = \dfrac{{3x}}{8} + \dfrac{{\sin 2x}}{4} + \dfrac{{\sin 4x}}{{32}} + c$

Note:
Integration is defined as the summation of all the discrete data. Differentiation is the opposite of the integration i.e. differentiation of the integration is equal to the value of the function or vice versa. Integration is also known as antiderivative. Here, we can make a mistake by just integrating the integrand without converting it to an integral form. This will give us the wrong answer. We need to remember the different integration formulas to simplify the expression.