## Introduction to the integral of sqrt(1-x^2)

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 an algebraic function √1-x^2. You will also understand how to compute √1-x^2 integral by using different integration techniques.

## What is the integration of √(1-x^2)?

The integral of √1-x^2 is an antiderivative of the √1-x^2 function which is equal to ½ arcsin x+½ x√1-x^2. It is also known as the reverse derivative of the function √1-x^2 which is an algebraic function. It can be calculated by using the power rule of integral. This rule is written as;

$\int x^n dx=\frac{x^{n+1}}{n+1}+c$

This formula says that the integral of any algebraic function with some exponent, can be calculated by adding 1 in its exponent and dividing by the new exponent i.e n+1.

### Integral of sqrt(1-x^2) formula

The formula of integral of square root x squared contains integral sign, coefficient of integration and the function as √x. It is denoted by ∫√1-x^2 dx. In mathematical form, the integral of √x is:

$int \sqrt{1-x^2} dx=\frac{1}{2}\sin^{-1}x+\frac{1}{2}x\sqrt{1-x^2}+c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. Use our integral calculator to integrate sqrt(1-x^2) online easily by using simple steps.

## How to calculate the integral sqrt(1-x^2)?

The sqrt(1-x^2) integral is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of a square root 1-x^2 by using:

- Trigonometric substitution method
- Integration by parts
- Definite integral

## Integral of square root 1-x^2 by using trigonometric substitution

The trigonometric substitution is a method of integration in calculus. It is used to calculate the integral of a function which is complex to be calculated by usual integration. Let’s discuss calculating the integration of under root 1-x^2 by using the trigonometric substitution calculator.

### Proof of sqrt(1-x^2) integral by using substitution method

To integrate square root (1-x^2) by using the substitution method, suppose that:

$I=\int\sqrt{1-x^2}dx$

In the trigonometric substitution, we substitute the variable with a trigonometric function. Now suppose that,

$x=\sin \theta$

and

$dx=\cos \theta d\theta$

Since,

$\cos \theta =\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}$

Using the new substitution in the above integral.

$I=\int \cos \theta .\cos\theta d\theta$

We can write it as;

$I=\int \cos^2\theta d\theta $

By using the following trigonometric formula,

$\cos^2 \theta=\frac{1+\cos 2\theta}{2}$

The integral will become,

$I=\int \frac{1+\cos 2\theta}{2}d\theta $

Separating the integrals,

$I=\int \frac{1}{2}d\theta +\int \frac{\cos 2\theta}{2}d\theta $

Integrating, each term,

$I=\frac{\theta}{2}+\frac{\sin 2\theta}{4}=\frac{\theta}{2}+\frac{2\sin \theta \cos \theta}{4}{2}nbsp;

$I=\frac{\theta}{2}+\frac{\sin \theta\cos \theta}{2}$

Writing cos in terms of sin .

$I=\frac{\theta}{2}+\frac{1}{2}\times \sin \theta\sqrt{1-\sin^2\theta}$

Now substituting the value of ,

$I=\frac{1}{2}\sin^{-1}x+\frac{1}{2}x\sqrt{1-x^2}+c$

Hence we have verified the integration of under root 1-x^2 by using the trigonometric substitution method. The u-substitution method is also another way to evaluate integrals.

**Integral sqrt(1-x^2) 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 sqrt(1-x^2) by using integration by parts.

**Proof of sqrt(1-x^2) integral by using integration by parts**

To integrate square root (1-x^2) using integration by parts, we can use the following formula:

$I=f(x)\int g(x)dx -\int [f’(x)\int g(x)dx]$

Suppose that,

$f(x) = \sqrt{1-x^2}$

$g(x) =1{2}nbsp;

Using these values in the above formula,

$I=\sqrt{1-x^2}\int 1dx-\int [\frac{d}{dx}(\sqrt{1-x^2})\int 1dx]dx$

Integrating,

$I=x\sqrt{1-x^2}-\int \left(\frac{-2x}{2\sqrt{1-x^2}}\times x\right)dx$

$I=x\sqrt{1-x^2}+\int \left(\frac{x}{\sqrt{1-x^2}}\times x\right)dx$

$I=x\sqrt{1-x^2}+\int \frac{x^2}{\sqrt{1-x^2}}$

$I=x\sqrt{1-x^2}+\int \frac{x^2-1+1}{\sqrt{1-x^2}}dx$

Separating the integrals,

$I=x\sqrt{1-x^2}+\int \frac{1}{\sqrt{1-x^2}}dx-\int \frac{1-x^2}{\sqrt{1-x^2}}dx$

Simplifying by rationalising 1-x^2/√1-x^2, we get

$I=x\sqrt{1-x^2}+\sin^{-1}x-\int \sqrt{1-x^2}dx$

Since,

$I=\int \sqrt{1-x^2}dx$

$I=x\sqrt{1-x^2}+\sin^{-1}x-I$

Or,

$2I=x\sqrt{1-x^2}+\sin^{-1}x$

$I=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x+c$

Hence the integral of sqrt(1-x^2) is ½ arcsin x+½ x√1-x^2, where c is a constant known as an integration constant.

## Frequently Asked Questions

### How do you integrate square √ 1 - x 2?

To calculate the integral sqrt(1-x^2), we can use different integration techniques, such as integration by parts rule and the trigonometric substitution method. Mathematically, the integral of 1-x2, is calculated as;

$\int \sqrt{1-x^2} dx=\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x+c$

### What is the integral of √ x?

The integration of square root x is equal to 2x^(3/2)/3. It is expressed as;

$\int \sqrt{x}dx=\frac{x^{½+1}}{1/2+1}=\frac{2x^{3/2}}{3}+c{2}nbsp;