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

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.

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.

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

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

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

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

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

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