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
\[\int {\frac{1}{{\sqrt {1 – {x^2}} }}dx} = – {\cos ^{ – 1}}x + c\]