Question

Evaluate the value of $\int {{{\cos }^4}} xdx$

Answer 1

The given function is indefinite since there is no limit given. The indefinite integral of a function is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.
In this question, we need to determine the indefinite integral of $\int {{{\cos }^4}} xdx$. For this we will use $\cos 2\theta = 2{\cos ^2}\theta - 1$ trigonometric identity and properties of the integration.

Complete step by step solution:
Let the given integral be $I$ such that:
$I = \int {{{\cos }^4}} xdx$
This function can be written as
$I = {\int {\left( {{{\cos }^2}x} \right)} ^2}dx - - (i)$
Now we know the trigonometric identity $\cos 2\theta = 2{\cos ^2}\theta - 1$
Hence by using this trigonometric equation in equation (i), we can write
$I = \int {{{\left( {\dfrac{{\cos 2x + 1}}{2}} \right)}^2}} dx$
This can be further written as
$I = \dfrac{1}{4}\int {\left( {{{\cos }^2}\left( {2x} \right) + 1 + 2\cos 2x} \right)} dx - - (ii)$
Now again, we use the trigonometric equation $\cos 2\theta = 2{\cos ^2}\theta - 1$ in equation (ii), hence we get
$I = \dfrac{1}{4}\int {\left( {\left( {\dfrac{{\cos 4x + 1}}{2}} \right) + 1 + 2\cos 2x} \right)} dx$
Taking LCM in the above equation, we get

$$I = \dfrac{1}{8}\int {\left( {\cos 4x + 1 + 2 + 4\cos 2x} \right)} dx \\\ = \dfrac{1}{8}\int {\left( {\cos 4x + 3 + 4\cos 2x} \right)} dx \\\$$

Now, splitting the terms and applying separate integration as:
$I = \dfrac{1}{8}\left[ {\int {\cos 4xdx + \int {3dx} + \int {4\cos 2xdx} } } \right]$
Now let’s integrate the above obtained equation; we get

$$I = \dfrac{1}{8}\left[ {\int {\cos 4xdx + \int {3dx} + \int {4\cos 2xdx} } } \right] \\\ = \dfrac{1}{8}\left[ {\dfrac{{\sin 4x}}{4} + 3x + 4\dfrac{{\sin 2x}}{2}} \right] + c \\\$$

{Where$\int {\cos \left( {ax} \right)dx = \dfrac{1}{a}\sin \left( {ax} \right)}$}
Hence by solving this obtained equation, we get
$I = \left[ {\dfrac{{\sin 4x}}{{32}} + \dfrac{{3x}}{8} + \dfrac{{\sin 2x}}{4}} \right] + c$
Hence we can say
$\int {{{\cos }^4}} xdx = \dfrac{{\sin 4x}}{{32}} + \dfrac{{3x}}{8} + \dfrac{{\sin 2x}}{4} + c$

Important equations used:
1. $\cos 2\theta = 2{\cos ^2}\theta - 1$
2. $\int {\cos \left( {ax} \right)dx = \dfrac{1}{a}\sin \left( {ax} \right)}$
3. $\int {dx = x + c}$

Note:
While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.