Integral of Cos^2(3X)

Integral of cos^2(3x) along with its formula and proof with examples. Also learn how to calculate integration of cos^2(3x) with step by step examples.

Alan Walker-

Published on 2023-04-14

Introduction to integral of cos^2(3x)

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

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

The integration of cos^2 3x is an antiderivative of the cos square 3x function which is equal to x/2 + sin (6x)/12. It is also known as the reverse derivative of the cos^2(3x) function which is a trigonometric identity. The sine function is the ratio of the opposite side to the hypotenuse of a triangle which is written as:

Sin = opposite side/hypotenuse

Integral of cos2 (3x)i formula

The formula of the integral of cos square (3x) contains the integral sign, coefficient of integration, and the function as cos. It is denoted by ∫{cos2 (3x)}dx. In mathematical form, the cos^2(3x) formula of integral is:

$\int \cos^2(3x)dx =\frac{x}{2} +\frac{\sin(6x)}{12} + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of the integral. The above formula can also be used to compute the integral of cos^4(2x).

How to integrate cos^2(3x) with respect to x?

The cos^2(3x) integral 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:

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

Integral of cos 3x squared 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 squared 3x by using integration by parts.

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

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

$I = \cos(3x).\cos(3x)$

To find the integral of cos square 2x, we can write the above expression as the product of cos(2x) with itself.

Applying the integral we get,

$I = \int[\cos(3x).\cos(3x)]dx$

Since the method of integration by parts is:

$\int [f(x).g(x)]dx = f(x)\int g(x)dx – \int [f’(x).\int g(x)dx]dx$

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

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

It can be written as:

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

Now by using a trigonometric identity sin2(3x) = 1- cos(6x)/2. Therefore, substituting the value of sin2 3x in the above equation, we get:

$I = \cos(3x).\frac{\sin(3x)}{3}+ \int \frac{1- \cos(6x)}{2}dx$

Integrating remaining terms,

$I = \cos(3x).\frac{\sin(3x)}{3} +\frac{x}{2} -\frac{sin(6x)}{12}$


$I = \frac{sin(6x)}{6}+\frac{x}{2} – \frac{sin(6x)}{12}$


$I = \frac{x}{2} + \frac{sin(6x)}{12}$

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

$\int \cos^2(3x)dx = \frac{x}{2} + \frac{sin(6x)}{12} + c{2}nbsp;

Similarly, we can use this method to find the integral of cos^3(2x).

Cos^2(3x) integral 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 cos squared by using the substitution method.

Proof of Integral of cos2(3x) by using substitution method

To integrate cos^2 (3x) dx by using substitution method, suppose that:

$I=\int \cos^2(3x)=\int [1- \sin^2(3x)]dx$

Further we can sin2(3x) can be substituted as sin2(3x) = 1 – cos(6x)/2. Then the above equation will become.

$I = x - \int\frac{[1- cos(6x)]}{2}dx$


$I = x –\frac{x}{2} +\frac{\sin(6x)}{12}$


$I = \frac{x}{2} +\frac{\sin(6x)}{12} + c$

Hence the integration of cos2(3x) is verified by using the substitution method. You can also use our u substitution formula calculator. It assists you to compute integrals by writing them in a simpler form.

Integral of cos2(3x) 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:

$\int^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integration of cos^2 3x by using the definite integral.

Proof of integral of cos2(3x) by using definite integral

To integrate cos^2(3x) by using a definite integral solver, we can use the interval from 0 to 2π or 0 to π. Let’s compute the integral of cos2(3x) from 0 to 2π.

The definite integral of cos^2(3x) can be written as:

$\int^{2\pi}_0 \cos^2(3x)dx = \left|\frac{x}{2} + \frac{\sin(6x)}{12}\right|^{2\pi}_0$

Substituting the value of the limit we get,

$\int^{2\pi}_0 \cos^2(3x)dx= \left[\frac{2π}{2}+\frac{\sin(6π)}{12}\right]-\left[0 +\frac{\sin 0}{12}\right]$

$\int^{2\pi}_0 \cos^2(3x)dx =\pi - 0$

Therefore, the integral of cos2 3x from 0 to 2π is

$\int^{2\pi}_0 \cos^2(3x)dx =\pi$

Which is the calculation of the definite integral of cos2(3x). Now to calculate the integration of cos2(3x) between the interval 0 to π, we just have to replace 2π with π. Therefore,

$\int^\pi_0 \cos^2(3x)dx = \left|\frac{x}{2}+\frac{\sin(6x)}{12}\right|^\pi_0$

$\int^\pi_0 \cos^2(3x)dx = \left[\frac{π}{2} +\frac{\sin(6π)}{12}\right] - \left[0 + \frac{\sin 0}{12}\right]$

$\int^\pi_0 \cos^2(3x)dx =\frac{π}{2}-\frac{0}{4}$

$\int^\pi_0 \cos^2(3x)dx = \frac{π}{2}$

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

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