Introduction to the integral of 1/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 1/1-x^2. You will also understand how to compute 1/1-x^2 integral by using different integration techniques.

What is the integral of 1 by 1-x^2?

The integral of 1/1-x^2 is an antiderivative of the 1/1-x^2 function which is equal to ½ ln(1-x/1+x). It is denoted as ∫1/1-x^2 dx= 1/2 ln(1-x/1+x). It is also known as the reverse derivative of the function 1/1-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 1/(1-x^2) formula

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

$\int \frac{1}{1-x^2}=\frac{1}{2}\ln(\frac{1-x}{1+x})+c$

Where c is any constant involved, dx is the coefficient of integration calculator and ∫ is the symbol of integral. 

How do you integrate 1/1-x^2?

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

1. Trigonometric substitution method
2. Integration by partial fraction

Integral of 1/1-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 integral of the 1/1-x^2 by using the trig-substitution calculator.

Proof of Integral of square root of x by using substitution method

To proof the integral of 1/sqrt(1-x^2) by using substitution method, suppose that:

$I=\int \frac{1}{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,

$\cos^2 \theta=1-\sin^2\theta$

Using the new substitution in the above integral.

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

We can write it as;

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

Integrating, 

$I=\ln|\sec \theta +\tan \theta|$

Now substituting the value of ,

$I=\ln|\sec (\sin^{-1})+\tan(\sin^{-1})|$

Now to simplify the above expression, we will draw a triangle, whose sides are AB, BC and CA.

By using the Pythagorean’s theorem for the above triangle,

$x^2+AB^2=1^2$

$AB^2=1-x^2$

$AB=\sqrt{1-x^2}$

Since x=sin , therefore,

$\tan \theta = \frac{BC}{AB}=\frac{x}{\sqrt{1-x^2}}$

$\sec \theta=\frac{AC}{AB}= \frac{1}{\sqrt{1-x^2}}$

Now using these values in the integral,

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

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

Integral of 1/1-x^2 by using integration by partial fraction

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 partial fraction is a method of solving the integral by breaking down the functions into different parts. Let’s discuss calculating the integral of 1/1-x^2 by using integration by partial fraction.

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

To integrate the function 1/1-x^2 using integration by partial fraction, we have to break down the integral into two different parts.

Suppose that, 

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

And,

$\frac{1}{1-x^2} = \frac{1}{(1-x)(1+x)}$

By partial fraction, 

$\frac{1}{1-x^2} =\frac{A}{1-x}+\frac{B}{1+x}$

After solving, the values of A and B are:

$A=\frac{1}{2}$

$B=\frac{-1}{2}$

By using these values, 

$\frac{1}{1-x^2} =\frac{1}{2(1-x)}+\frac{-1}{2(1+x)}$

Substituting it in the original integral,

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

Now, 

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

Integrating, 

$I=\frac{1}{2}\left(\ln(1-x)-\ln(1+x)\right)$

Or, 

$I=\frac{1}{2}\ln\left(\frac{1-x}{1+x}\right)+c$

Hence the integral of 1/sqrt(1-x^2) is ½ln(1-x/1+x) +c, where c is a constant known as an integration constant. You can use our integration by partial fraction calculator to calculate integral more easily.

Frequently Asked Questions

How do you integrate square √ 1 - x 2?

To calculate the integral of 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 1/√ x?

The integration 1/sqrt x is equal to 2x. It is expressed as;

$\int \frac{1}{\sqrt{x}}dx=\frac{x^{-½+1}}{-½+1}=\frac{x^{½}}{½}+c$