## Introduction to integral of cos^4(2x)

In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.

Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function cos^4(2x). You will also understand how to compute cos^4(2x) integral by using different integration techniques.

## What is the integral of cos^4(2x)?

The antiderivative of the fourth power of the cosine of twice a certain variable, denoted by cos^4(2x), can be expressed as a function of the sine, namely 3x/8 + sin(8x)/64 + sin(4x)/8+c. This is also referred to as the inverse derivative of the sine function, which is a trigonometric identity.

The cosine function represents the ratio of the adjacent side to the hypotenuse of a triangle, symbolized as:

cos = adjacent side / hypotenuse

### Integral of cos4 2x formula

The formula of integral of cosin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(cos^4 2x)dx. In mathematical form, the integral of sin4x is:

$∫\cos^4(2x)dx=\frac{3x}{8}+\frac{1}{8}\sin4x+\frac{1}{64}\sin8x + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. The above formula can also be used to calculate the integration of cos^2(2x).

## How to calculate the integral of cos^{4}(2x)?

The integral of cos^4(2x) is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using:

- Integration by parts
- Substitution method
- Definite integral

## Integral of cos^{4}(2x) by using integration by parts

In calculus, the derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving integral of two functions combined together. Let’s discuss calculating the integral of cos^4(2x) by using integration by parts.

### Proof of integral of cos^4(2x) by using integration by parts

Since we know that the function cosine x can be written as the product of two functions. Therefore, we can calculate the integral of sin^4x by using integration by parts. For this, suppose that:

$I = \cos^4(2x)=\cos^3(2x).\cos(2x)$

Applying the integral we get,

$I = ∫\cos^3(2x).\cos(2x)dx$

Since the integral by parts formula is:

$∫[f(x).g(x)]dx = f(x).∫g(x)dx - ∫[f’(x).∫g(x)]dx$

Now replacing f(x) and g(x) by cos x, we get,

$I = \cos^3(2x).\frac{\sin(2x)}{2} + ∫\left[3\cos^2(2x).2\sin(2x).\frac{\sin(2x)}{2}\right]dx$

It can be written as:

$I = \cos^3(2x).\frac{\sin(2x)}{2} + 3∫[\sin^2(2x).\cos^2(2x)]dx$

Now by using a trigonometric identity 4sin2xcos2x = sin^{2}(2x). Therefore, substituting the value of sin^{2}2x in the above equation, we get:

$I=\cos^3(2x).\frac{\sin(2x)}{2}+\frac{3}{4}∫[\sin^2(4x)]dx$

Now again using trigonometric formula as, Sin^{2}(4x)= 1 – cos8x/2 then,

$I=\cos^3(2x).\frac{\sin(2x)}{2}+\frac{3}{4}∫\left(\frac{1–\cos8x}{2}\right)dx$

Integrating remaining terms, we get

$I=\cos^3(2x).\frac{\sin(2x)}{2}+\frac{3x}{8}– \frac{3\sin4x}{64}+ c$

This method of solving integrals can be used to integrate cos^2(3x).

## Integral of cos^{4}(2x) by using substitution method

The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of cos4 (2x) by using the substitution method.

### Proof of Integral of cos^4(2x) by using substitution method

To prove the integral of cos^4(2x) by using substitution method, suppose that:

$I = ∫\cos^4(2x)dx$

Suppose that we can write the above integral as:

$I=\int \cos^2(2x)\cos^2(2x)dx$

By using trigonometric identities, we can write the above equation by using 2cos^{2}(2x) = 1 + cos4x, therefore,

$I=\frac{1}{4}\int \left(\frac{1+\cos(4x)}{2}\right)^2dx$

Simplifying,

$I = \frac{1}{4}∫(1+\cos^2(4x) + 2\cos(4x))dx$

Now to simplify more, we can use another formula as:

$cos^2(4x) = \frac{1+\cos8x}{2}$

Now, using this formula in the integral,

$I=\frac{1}{4}∫\left(1+(1+cos8x)/2 + 2cos4x\right)dx$

Moreover,

$I = \frac{1}{4}∫1dx+∫\frac{1}{8}dx+∫\frac{\cos8x}{8}+2∫\cos4xdx$

Integrating each term,

$I =\frac{3x}{8}+\frac{\sin8x}{64}+\frac{\sin2x}{8}$

Hence the integration of cos^4(2x) is verified by using substitution method. You can also use our u-substitution calculator to verify the above calculations.

## Integral of cos^4(2x) by using definite integral

The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of cos^4(2x) by using the definite integral.

### Proof of integral of cos^4(2x) by using definite integral

To compute the integral of cos^4(2x) by using a definite integral calculator, we can use the interval from 0 to 2π or 0 to π. Let’s compute the integral of sin^3x from 0 to 2π.

The indefinite integral of cos^4(2x) can be written as:

$∫^{2\pi}_0 \cos^4(2x)dx = \left|\frac{3x}{8} + \frac{\sin(8x)}{64}+\frac{\sin(4x)}{8}]\right|^{2π}_0$

Substituting the value of limit we get,

$∫^{2π}_0 \cos^4(2x)dx =\left[\frac{6π}{8}+\frac{sin8(2π)}{64} +\frac{\sin4(2π)}{8}\right] – [0 +sin 0 + sin 0]$

Therefore, the integral of cos4x from 0 to 2π is

$∫^{2π}_0 \cos^4(2x)dx = \frac{6π}{8}$

Which is the calculation of the definite integral of cos^4(2x). Now to calculate the integral of cos^4(2x) between the interval 0 to π, we just have to replace 2π by π. Therefore,

$∫^π_0 \cos^4(2x)dx=\left|\frac{3x}{8} + \frac{\sin8x}{64} + \frac{\sin4x}{8}\right|^\pi_0$

$∫^\pi_0 \cos^4(2x)dx=\left[\frac{3π}{8}+\frac{\sin8π}{64}+\frac{\sin4π}{8}\right] – 0$

$∫^π_0 \cos^4(2x)dx =\frac{3π}{8}$

Therefore, the integral of cos^{4}(2x) from 0 to π is 3π/8.