# Integral Of Sin Cube X

Integral of sin cube x along with its formula and proof with examples. Also learn how to calculate integration of sin cube x with step by step examples.

Alan Walker-

Published on 2023-04-13

## Introduction to integral of sin^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 solver 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 sin cube integral by using different integration techniques.

## What is the integral of sin^3x?

The integral of sin^3x is an antiderivative of sine function which is equal to cos3x/3 – cos x + 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:

Sin = opposite side / hypotenuse

### Integral of sin^3x formula

The formula of integral of sin3 x contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin3x)dx. In mathematical form, the integral of sin^3x is:

$∫\sin^3xdx=\frac{\cos3x}{3}– \cos x + c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. Similarly, the integral sin square is equal to x/2 - (sin2x)/4 + c.

## How to calculate the integral of sin^3x?

The integral of sin^3x 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 sin x 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 integral of two functions combined together. Let’s discuss calculating the integral of sin squared x by using by parts integral calculator.

### Proof of integral of sin^3x by using integration by parts

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

$I=\sin^3x=\sin x.\sin^2x$

Applying the integral we get,

$I=∫\sin x.\sin^2xdx$

Since the method of integration by parts is:

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

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

$I=-\sin^2x.\cos x + ∫[2\sin x\cos x.\cos x]dx$

It can be written as:

$I= -\sin^2x.\cos x + 2∫[\sin x.\cos^2x]dx$

Now by using a trigonometric identity cos2x = 1 – sin2x. Therefore, substituting the value of cos2x in the above equation, we get:

$I=-\sin^2x.\cos x+2∫\sin x(1 – \sin^2x)dx$

Integrating remaining terms,

$I=-\sin^2x.\cos x-2\cos x – 2\int\sin^3x$

Since we know that,

$I=\int \sin^3xdx$

$I= -\sin^2x.\cos x -2\cos x – 2I$

$I=-(1-\cos^2x)\cos x -2\cos x-2I$

$I=-\cos x+\cos^3x-2\cos x-2I$

Or,

$3I =\cos^3x -3cos x$

Hence the integral of sin^3x is equal to,

$∫\sin^3xdx =\frac{\cos^3x}{3}-\cos x+c$

Also, you can find the integral of sin(e^x) by using the integration by parts formula.

## Integral of sin^3x 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 cube by using the substitution method.

### Proof of Integral of sin^3x by using substitution method

To proof the integral of sin^3x by using substitution method, suppose that:

$I = ∫\sin^3xdx$

Suppose that we can write the above integral as:

$I = ∫\sin x.\sin^2xdx$

By using trigonometric identities, we can write the above equation by using sin2x = 1 – cos2x, therefore,

$I = ∫\sin x.( 1 – \cos^2x)dx$

Simplifying,

$I= ∫(\sin x – \sin x\cos^2x)dx$

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

$I_1 = ∫\sin x.dx = - \cos x$

Now to evaluate second integral,

$I_2 = -∫\sin x\cos^2xdx$

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

$I_2 = ∫u^2du$

Integrating with respect to u.

$I_2 = \frac{u^3}{3}$

Substituting the value of u we get,

$I_2 =\frac{\cos^3x}{3}$

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

$I = -\cos x+ \frac{\cos^3x}{3}+c$

Hence the integration of sin^3x is verified by using substitution method. It can be also verified by using the trigonometric substitution

## Integral of sin^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:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of sin^3x by using the indefinite integral.

### Proof of integral of sin^3x by using definite integral

To compute the integral of sin^3x by using a definite integral, 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 sin^3x can be written as:

$∫^2π_0 \sin^3x dx=\left|\frac{\cos^3x}{3} –\cos x\right|^{2π}_0$

Substituting the value of limit we get,

$∫^{2π}_0 \sin^3x dx=\frac{\cos^3 2π}{3} -\frac{\cos^3 0}{3} –[\cos 2π - \cos 0]$

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

$∫^{2π}_0 \sin^3x dx=-\frac{1}{2}$

Which is the calculation of the definite integral solver of sin^3x. Now to calculate the integral of sinx between the interval 0 to π, we just have to replace π by π. Therefore,

$∫^π_0 \sin^3xdx=\left|\frac{\cos^3x}{3} –\cos x\right|^π_0$

$∫^π_0 \sin^3x dx=\left[\frac{\cos^3π}{3}-\frac{\cos^3 0}{3}\right]–[\cos π-\cos 0]$

$∫^π_0 \sin^3xdx=\frac{1}{2}–1 – 1 +1$

$∫^π_0 \sin^3xdx=-\frac{1}{2}$

Therefore, the integral of sin^3x from 0 to π is -1/2.