# $\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}\,}dx$ Rule

## Formula

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

The indefinite integral of one by square root of one minus square of a variable is equal to the sum of the inverse sine function and the constant of integration.

### Introduction

When a variable is denoted by $x$, the multiplicative inverse (or reciprocal) of the square root of the subtraction of the square of variable $x$ from one is written in mathematical form as follows.

$\dfrac{1}{\sqrt{1-x^2}}$

The indefinite integration for this irrational algebraic function in reciprocal form is written in the following mathematical form in calculus.

$\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}\,\,}dx$

According to the inverse trigonometry, the inverse sine function is written as $\sin^{-1}{(x)}$ or $\arcsin{(x)}$ in mathematics. If the constant of integration is denoted by $c$, then the indefinite integration of this irrational function is equal to the sum of the inverse sine function and the constant of integration.

$(1).\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}\,\,}dx$ $\,=\,$ $\sin^{-1}{(x)}+c$

$(2).\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{1-x^2}}\,\,}dx$ $\,=\,$ $\arcsin{(x)}+c$

It is used as a formula in calculus for evaluating the integrals of the functions.

##### Other forms

This integration rule can be expressed in terms of any variable in the integral mathematics as follows.

$(1).\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{1-b^2}}\,\,}db$ $\,=\,$ $\sin^{-1}{(b)}+c$

$(2).\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{1-t^2}}\,\,}dt$ $\,=\,$ $\arcsin{(t)}+c$

$(3).\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{1-y^2}}\,\,}dy$ $\,=\,$ $\sin^{-1}{(y)}+c$

#### Proofs

There are two possible mathematical approaches for proving this integral rule in mathematics.

##### Calculus method

Learn how to derive the integration for the multiplicative inverse of the square root one minus square of a variable in calculus method.

##### Trigonometric method

Learn how to derive the integration for the reciprocal of the square root of the subtraction of the variable squared from one in trigonometric method.

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