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