# Integral Of Sin^5x

Integral of sin^5x along with its formula and proof with examples. Also learn how to calculate integration of sin^5x with step by step examples.

Alan Walker-

Published on 2023-04-14

## Introduction to integral of sin^5x

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 sin^5x integral by using different integration techniques.

## What is the sin^5x integration?

The integral of sin^5x is an antiderivative of sine function which is equal to –cos5x/5 + 2cos3x/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

### Integration sin^5x formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin5x)dx. In mathematical form, the integral of sin5x is:

$∫\sin^5xdx = –\frac{\cos5x}{5}+2\frac{\cos^3x}{3}–\cos x +c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. If we reduce the power of sin^5x by 1, we can compute the integral of sin^4x.

## How to calculate the integral sin^5x?

The sin^5x 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. Trigonometric formulas
2. Substitution method
3. Definite integral

## Integral of sin^5x by using Trigonometric formulas

The substitution method also involves trigonometric formulas that helps to solve integrals easily. Let’s understand how to integrate sin^5x by using different trigonometric formulas. Also, learn to calculate the integral of sin square x.

### Proof of integral of sin^5x by using Trigonometric formulas

To prove the sin^5x integration, we use different trigonometric formulas. Therefore, suppose that

$I = \sin^5x = \sin^3x.\sin^2x$

To find the integral of sin cube, you can write the above integral as the product of two functions sin x and sin^2x. Now applying the integral we get,

$I = ∫\sin^3x.\sin^2xdx$

Since sin2x = 1 – cos2x

$I = ∫\sin^3x[1 – \cos^2x]dx$

Now,

$I = ∫[\sin^3x –\sin^3x.\cos^2x]dx$

Moreover,

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

$I = ∫[\sin x.(1 – \cos^2x) –\sin^3x.\cos^2x]dx$

Again,

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

Now separating integrals,

$I = ∫\sin xdx – ∫\sin x.\cos^2xdx –∫\sin^3x.\cos^2xdx$

Suppose that cos x = t and dt = - sin x, then

$I = ∫\sin xdx – ∫t^2.(-dt) –∫\sin^2x\sin x.t^2dx$

Again,

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

The integral will become after substituting the value of t,

$I=∫\sin xdx+∫t^2.dt+∫(1 – t^2).t^2dt$

And,

$I=∫\sin xdx+∫t^2.dt+∫(t^2 – t^4)dt$

Integrating,

$I=-\cos x+\frac{t^3}{3}+\frac{t^3}{3}–\frac{t^5}{5}$

Substituting the value of t,

$I=-\cos x+\frac{2\cos^3x}{3}–\frac{\cos^5x}{5} + c$

Hence we have verified the integral sin^5 by using trigonometric formulas.

## Integral of sin^5x 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 power 5 by using the substitution method.

### Proof of sin^5x integration by using substitution method

To proof the integral of sin^5x by using substitution method, suppose that:

$I = ∫\sin^5xdx$

Suppose that we can write the above integral as:

$I = ∫\sin^4x.\sin xdx$

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

$I = ∫(1 - \cos^2x)^2\sin xdx$

Now suppose that,

$u = \cos x\quad\text{and}\quad du=-\sin xdx$

$I=-∫(1 – u^2)^2du$

Now to simplify more,

$I=-∫(1 + u^4 – 2u^2)du$

Now integrating each term with respect to u,

$I=-u–\frac{u^5}{5}+\frac{2u^3}{3}+c$

Now substituting the value of u back here,

$I=-\cos x–\frac{\cos^5x}{5}+\frac{2\cos^3x}{3}+c$

Hence the integration of sin^5x is verified by using u-substitution formula calculator.

## Integral of sin^5 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^5x by using the definite integral.

### Proof of integral of sin^5x by using definite integral

To integrate sin^5x 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 definite integral of sin^5x can be written as:

$∫^{2π}_0 \sin^5xdx = -\cos x+\frac{2\cos^3π}{3}–\frac{\cos^5x}{5}|^{2π}_0$

Substituting the value of limit we get,

$∫^{2π}_0 \sin^5x dx =\left[-\cos π + \frac{2\cos^3π}{3}–\frac{\cos^5π}{5}\right] –\left[-\cos 0 + \frac{2\cos^30}{3}–\frac{\cos^50}{5}\right]$

Therefore, the sin^5x integral from 0 to 2π is

$∫^{2π}_0 \sin^5x dx=-1+\frac{2}{3}-\frac{1}{5}–\left(1 + \frac{2}{3}-\frac{1}{5}\right)$

$∫^{2π}_0 \sin^5x dx = 1–\frac{2}{3}+\frac{1}{5}–1–\frac{2}{3}+\frac{1}{5}$

$∫^{2π}_0 \sin^5x dx = -\frac{16}{15}$

Which is the calculation of the definite integral of sin^5x. Now to calculate the sin^5x integration between the interval 0 to π, we just have to replace 2π by π. Therefore,

$∫^π_0 \sin^5x dx = \left|-\cos x +\frac{2\cos^3π}{3}–\frac{\cos^5x}{5}\right|^π_0$

$∫^π_0 \sin^5x dx =\left[- \cos \frac{π}{2} + \frac{2\cos^3 \frac{π}{2}}{3} – \frac{\cos^5\frac{π}{2}}{5}\right]–\left[-\cos 0 + \frac{2\cos^30}{3} – \frac{\cos^50}{5}\right]$

$∫^π_0 \sin^5x dx = - 1 +\frac{2}{3} – \frac{1}{5} = -\frac{8}{15}$

Therefore, the integral sin^5x from 0 to π is -8/15.