Introduction integral of cos^3(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 cubic sine. You will also understand how to compute cos^3(2x) integral by using different integration techniques.

What is the integral of cos^3(2x)?

The integral of cos^3(2x) is an antiderivative of sine function which is equal to sin 2x/2 –(1/6)sin3 2x + c. It is also known as the reverse derivative of sine function which is a trigonometric identity. The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as:

cos = adjacent side/hypotenuse

The integral of cos cube 2x is a common integral in calculus. It is useful to solve many integral problems such as the integral of cos^2(3x). 

Integral of cos^3(2x) formula

The formula of the integral of sin contains the integral sign, coefficient of integration, and the function as cosine. It is denoted by ∫(cos3 2x)dx. In mathematical form, the integral of sin^3x is:

$\int \cos^3(2x)dx = \frac{\sin(2x)}{2} –\frac{\sin^3(2x)}{6} + c{2}lt;/p>

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. In the above integration formula, replacing cos^3(2x) by cos^2(2x) will give the integration of cos^2(2x).

How to calculate the integral of cos cube (2x)?

The integral cos^3 2x dx is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate the integral of cosine by using:

1. Integration by parts
2. Substitution method
3. Definite integral

Integral of cos 2x cubic by using integration by parts

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 the integral of two functions combined together. Let’s discuss calculating the integral of cos cubic power x by using integration by parts.

Proof of integral of cos^3(2x) by using integration by parts

Since we know that the function cosine cube 2x can be written as the product of two functions. Therefore, we can integrate cos^3 (2x) dx by using integration by parts. For this, suppose that:

$I = \cos^3(2x) = \cos(2x). \cos^2(2x){2}lt;/p>

Applying the integral we get,

$I = \int (\cos(2x).\cos^2(2x))dx{2}lt;/p>

Since the method of integration by parts is:

$\int [f(x).g(x)] = f(x).\int g(x)dx - \int [f’(x).\int g(x)]dx{2}lt;/p>

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

$I = \cos^2 (2x).\frac{\sin (2x)}{2} + \int [2\cos (2x)\sin (2x).\frac{\sin (2x)}{2}]dx{2}lt;/p>

It can be written as:

$I = \cos^2 (2x).\frac{\sin (2x)}{2} + \int [\sin^2(2x).\cos(2x)]dx{2}lt;/p>

Now by using a trigonometric identity sin22x = 1 – cos2 2x. Therefore, substituting the value of sin2 2x in the above equation, we get:

$I = \cos^2(2x).\frac{\sin (2x)}{2} + \int \cos(2x)[1 – \cos^2(2x)]dx{2}lt;/p>

Integrating remaining terms,

$I = \cos^2(2x).\sin(2x) +\frac {\sin(2x)}{2} – \int \cos^3(2x)dx{2}lt;/p>

Since we know that I = ∫cos³(2x)dx

$I = \cos^2(2x).\frac{\sin (2x)}{2} +\frac{\sin (2x)}{2} – I{2}lt;/p>

Or,

$2I = \cos^2(2x).\frac{\sin (2x)}{2} + \frac{\sin (2x)}{2}{2}lt;/p>

For more simplification, substitute cos²x = 1 – sin²x

$2I = \frac{\sin(2x)(1 – \sin^2(2x))}{2} + \frac{\sin(2x)}{2}{2}lt;/p>

Now,

$2I = \frac{\sin 2x}{2} – \frac{sin^3(2x)}{2} + \frac{sin(2x)}{2}{2}lt;/p>

$2I = \sin(2x) – \frac{\sin^3(2x)}{2}{2}lt;/p>

Now dividing by 3 on both sides,

$I = sin 2x – \frac{\sin^3(2x)}{4} + c{2}lt;/p>

Hence the cos^3(2x) integral is equal to,

$\int \cos^3 (2x)dx = \sin (2x) – \frac{\sin^3(2x)}{4} + c{2}lt;/p> 

This method is also applicable to calculate an integral of cos square ax.

Integral of cos^3(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 sin squared by using the substitution method.

Proof of integral cos^3 2x dx by using substitution method

The method of substitution involves different types of substitution,  trigonometric and u-substitution. You can also use the u-substitution calculator to solve integrals. To prove the integral of cos³(2x) by using the substitution method, suppose that:

$I = \int \cos^3(2x)dx{2}lt;/p>

Suppose that we can write the above integral as:

$I = \int [\cos(2x).\cos^2(2x)]dx{2}lt;/p>

By using trigonometric identities, we can write the above equation by using cos³(2x) = 1 – sin²(2x), therefore,

$I = \int [\cos(2x).( 1 – sin^2(2x)]dx{2}lt;/p>

Simplifying,

$I = \int [\cos(2x) – \cos(2x)\sin^2(2x)]dx{2}lt;/p>

Now to evaluate first integral, we will use the following steps,

$I_1 = \int \cos(2x).dx = \frac{sin(2x)}{2}{2}lt;/p>

Now to evaluate second integral,

$I_2 = -\int \cos(2x).\sin^2(2x) dx{2}lt;/p>

Suppose that u = sin 2x and du = 2cos 2x dx, then

$I_2 = -\frac{1}{2}\int u^2 du{2}lt;/p>

Integrating with respect to u.

$I_2 = -\frac{u^3}{6}{2}lt;/p>

Substituting the value of u we get,

$I_2 = -\frac{\sin^3(2x)}{3}{2}lt;/p>

Now, using the value of the first and second integral in the above equation to get the final value of the integral.

$I = \frac{\sin(2x)}{2} – \frac{\sin^3(2x)}{6} + c{2}lt;/p>

Hence the integration of cos³(2x) is verified by using the substitution method. The trigonometric substitution calculator with steps also provides you an easy way to evaluate integrals by using trigonometric formulas.

Integral of cos³(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:

$\int^b_a f(x) dx = F(b) – F(a){2}lt;/p>

Let’s understand the verification of the integral of sin^2x by using the definite integral.

Proof of integral of cos³(2x) by using definite integral

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

$\int^{\frac{\pi}{4}}_0 \cos^3(2x) dx = \left|\frac{\sin(2x)}{2} – \frac{sin^3(2x)}{6}\right|^{\frac{\pi}{4}}_0{2}lt;/p>

Substituting the value of limit we get,

$\int^{\frac{\pi}{4}}_0 \cos^3(2x) dx = \left[\frac{\sin 2π/4}{2} – \frac{\sin^3 2π/4}{6}\right] – \left[sin 0 – \frac{sin^3 0}{3}\right]{2}lt;/p>

$\int^{\frac{\pi}{4}}_0 \cos^3(2x) dx = \frac{1}{2} – \frac{1}{6} = \frac{1}{3}{2}lt;/p>

Therefore, the integral of cos3x from 0 to π/2 is

$\int^{\frac{\pi}{4}}_0 \cos^3(2x) dx = \frac{1}{3}{2}lt;/p>

Which is the calculation of the definite integral of cos³(2x). Now to calculate the integral of cos cube 2x between the interval 0 to π, we just have to replace π/4 by π. Therefore,

$\int^\pi_0 \cos^3(2x)dx = \left|\frac{\sin(2x)}{2} – \frac{\sin^3(2x)}{6}\right|^\pi_0{2}lt;/p>

$\int^\pi_0 \cos^3(2x)dx = \left[\frac{\sin π}{2} – \frac{\sin^3(2π)}{6}\right] – \left[\frac{\sin 0}{2} – \frac{\sin^30}{3}\right]{2}lt;/p>

$\int^\pi_0 \cos^3(2x)dx = 0 – 0{2}lt;/p>

$\int^pi_0 \cos^3(2x)dx = 0{2}lt;/p>

Therefore, the cos^3(2x) integral from 0 to π is 0.