## Introduction to integral of sin^4x

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 sin4 x integral by using different integration techniques.

## What is the integral of sin^4x?

The integral of sin^4x is an antiderivative of sine function which is equal to 3x/8 – ¼.sin(2x) + 1/32.sin(4x)+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^{4}x formula

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

$∫\sin^4xdx=\frac{3x}{8}–\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}+c$

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

## How to calculate the integral of sin^4x?

The integral of sin^4x 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 sin^4x 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^4x by using integration by parts calculator.

### Proof of integral of sin^4x 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^2x by using integration by parts. For this, suppose that:

$I = \sin^4x = \sin^3x.\sin x$

Applying the integral we get,

$I=∫\sin^3x\sin xdx$

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,

$I=-\sin^3x.\cos x-∫[3\sin^2x\cos x.\cos x]dx$

It can be written as:

$I=-\sin^3x.\cos x + 3∫[\sin^2x.\cos^2x]dx$

Now by using a trigonometric identity 4sin^{2}xcos^{2}x = sin^{2}2x. Therefore, substituting the value of sin2 2x in the above equation, we get:

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

Now again using trigonometric formula as, Sin^{2}2x = 1 – cos4x/2 then,

$I=-\sin^3x.\cos x+\frac{3}{4}∫\left(\frac{1 –\cos4x}{2}\right)dx$

Integrating remaining terms,

$I=-\sin^3 x.\cos x+\frac{3x}{4}–\frac{3\sin4x}{32} + c$

Thus, the method of integrating by parts is suitable to calculate integrals of a product of two functions or a function with some power. For example, the integral of sin^3x is a function with exponent 3.

## Integral of sin^4x 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. Also learn how to integrate the integral of sin square theta.

### Proof of Integral of sin^4x by using substitution method

To proof the integral of sin^4x by using substitution method, suppose that:

$I = ∫\sin^4xdx$

Suppose that we can write the above integral as:

$I=∫\sin^2x.\sin^2xdx$

By using trigonometric identities, we can write the above equation by using sin^{2}x = 1 – cos2x/2, therefore,

$I = \frac{1}{4}∫(1 – \cos2x)^2dx$

Simplifying,

$I=\frac{1}{4}∫(1+\cos^22x – 2\cos2x)dx$

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

$\cos^22x = \frac{1+\cos4x}{2}$

Now, using this formula in the integral,

$I=\frac{1}{4}∫\left(1+\frac{(1+\cos4x)}{2}–2\cos2x\right)dx$

Moreover,

$I=\frac{1}{4}∫1dx+∫\frac{1}{8}dx+∫\frac{\cos4x}{8}–∫\frac{\cos2x}{2}dx$

Integrating each term,

$I =\frac{3x}{8}+\frac{\sin4x}{32}–\frac{\sin2x}{4}$

Hence the integration of sin^4x is verified by using substitution method. You can also use our u-substitution calculator that provides a step-by-step integration of any function.

## Integral of sin^4x 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^4x by using the definite integral.

### Proof of integral of sin^4x by using definite integral

To compute the integral of sin^4x 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^4x can be written as:

$∫^{2π}_0 \sin^4xdx=\left|\frac{3x}{8}+\frac{\sin4x}{32}–\frac{\sin2x}{4}\right|^{2π}_0$

Substituting the value of limit we get,

$∫^{2π}_0 \sin^4xdx=\left[\frac{6π}{8}+\frac{\sin8π}{32} – \frac{\sin4π}{4}\right]-\left[0 +\sin 0 – \sin 0\right]$

Therefore, the integral of sin4x from 0 to 2π is

$∫^{2π}_0 \sin^4x dx = \frac{3π}{4}$

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

$∫^π_0 \sin^4xdx=\left|\frac{3x}{8}+\frac{\sin4x}{32}–\frac{\sin2x}{4}\right|^{2π}_0$

$∫^π_0 \sin^4xdx=\left[\frac{3π}{8}+\frac{\sin4π}{32}–\frac{\sin2π}{4}\right] – 0$

$∫^π_0 \sin^4x dx = \frac{3π}{8}$

Therefore, the integral of sin4x from 0 to π is 3π/8.