Question

How do you find the antiderivative of ${\cos ^4}x$?

Answer 1

Here we must know that derivative means we need to find the differentiation and in antiderivative we need to do the integration. So in this from we need to integrate the term and get the result by using the formula of the trigonometric function ${\cos ^4}x$

Complete step by step solution:
Here we are given to find the antiderivative of the term ${\cos ^4}x$ which means we need to integrate the term ${\cos ^4}x$
So we must know that whenever we have the power to the trigonometric function and we need to integrate, then we need to convert it into the trigonometric function which has no power. Hence we need to apply the formula according to it.
So we know that $\cos 2x = 2{\cos ^2}x - 1$
So we write that
$\cos 2x = 2{\cos ^2}x - 1 \\\ {\cos ^2}x = \dfrac{{1 + \cos 2x}}{2} \\\$
So now we can write the integration to be done according to its symbol as:
$\int {{{\cos }^4}x.dx}$
So we can write this as:
${\int {\left( {{{\cos }^2}x} \right)} ^2}dx$
Now we need to substitute the value of ${\cos ^2}x$ from above in the above integration, we will get this as:
${\int {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)} ^2}dx$
Now we know that ${(a + b)^2} = {a^2} + {b^2} + 2ab$
So we can apply the above formula in the integration we have obtained.
So we will get:
${\int {\left( {\dfrac{{1 + \cos 2x}}{2}} \right)} ^2}dx$
$\int {\left( {\dfrac{{{1^2} + {{\cos }^2}2x + 2\cos 2x}}{4}} \right)} dx$
As we can take the constant out of the integration so we can write this integration as:
$\dfrac{1}{4}\int {\left( {1 + {{\cos }^2}2x + 2\cos 2x} \right)} dx$
Again we have the power in the term $\cos 2x$ so again applying the same formula and we will get:
$\dfrac{1}{4}\int {\left( {1 + \dfrac{{1 + \cos 4x}}{2} + 2\cos 2x} \right)} dx$
Taking LCM as $2$ we will get:

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

Now taking $2$ outside we will get:
$\dfrac{1}{8}\int {\left( {3 + \cos 4x + 4\cos 2x} \right)} dx$
Now we know that we can separate all the integration when they are in addition or subtraction so we can write it as:
$\int {\dfrac{3}{8}dx + } \int {\dfrac{{\cos 4x}}{8}dx + \int {\dfrac{{\cos 2x}}{2}} } dx$
Now we know that:
$\int {\left( {\cos nx} \right)} dx = \dfrac{{\sin nx}}{n}$
So we can write it as:
$\int {\dfrac{3}{8}dx + } \int {\dfrac{{\cos 4x}}{8}dx + \int {\dfrac{{\cos 2x}}{2}} } dx$

$$\dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{8.4}} + \dfrac{{\sin 2x}}{{2.2}} + c \\\ \dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} + \dfrac{{\sin 2x}}{4} + c \\\$$

Hence we get our result as $\dfrac{{3x}}{8} + \dfrac{{\sin 4x}}{{32}} + \dfrac{{\sin 2x}}{4} + c$

Note:
Here in these types of problems where we are asked to find the antiderivative, we must know that it means integration. We must keep in mind all the basic formulas of the integration and also the trigonometric formula. Even though we know all the formulas we must have the ability to know which formula is to be applied and that comes with the practice.