# Integral of 1 Over the Square Root of 1-x^2

In this tutorial we shall look at another important integration: the integral of 1 over the square root of ${1-x^2}$. This integral also uses an important formula.

The integral of 1 over the square root of ${1-x^2}$ is of the form
$\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx = } {\sin ^{ – 1}}x + c$

To prove this formula, consider
$\frac{d}{{dx}}\left[ {{{\sin }^{ – 1}}x + c} \right] = \frac{d}{{dx}}{\sin ^{ – 1}}x + \frac{d}{{dx}}c$

Using the derivative formula $\frac{d}{{dx}}{\sin ^{ – 1}}x = \frac{1}{{\sqrt {1 – {x^2}} }}$, we have
$\begin{gathered} \frac{d}{{dx}}\left[ {{{\sin }^{ – 1}}x + c} \right] = \frac{1}{{\sqrt {1 – {x^2}} }} + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {{{\sin }^{ – 1}}x + c} \right] = \frac{1}{{\sqrt {1 – {x^2}} }} \\ \Rightarrow \frac{1}{{\sqrt {1 – {x^2}} }} = \frac{d}{{dx}}\left[ {{{\sin }^{ – 1}}x + c} \right] \\ \Rightarrow \frac{1}{{\sqrt {1 – {x^2}} }}dx = d\left[ {{{\sin }^{ – 1}}x + c} \right]\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

Integrating both sides of equation (i) with respect to $x$, we have
$\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} = \int {d\left[ {{{\sin }^{ – 1}}x + c} \right]}$

As we know that by definition the integration is the inverse process of the derivative, so we have
$\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} = {\sin ^{ – 1}}x + c$

We know that the derivative of $\frac{d}{{dx}}{\cos ^{ – 1}}x = – \frac{1}{{\sqrt {1 – {x^2}} }}$

This formula can also be written as
$\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} = – {\cos ^{ – 1}}x + c$