# Integration of 1 Over the Square Root of (x^2+a^2)

In this tutorial we shall discuss the integration of 1 over the square root of x^2+a^2, and this is another important form of integration.

The integration of $\frac{1}{{\sqrt {{x^2} + {a^2}} }}$ is of the form
$\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx = } {\sin ^{ – 1}}\left( {\frac{x}{a}} \right) + c$

To prove this formula, putting $x = a\sinh t$ we have $dx = a\cosh tdt$, $t = {\sin ^{ – 1}}\left( {\frac{x}{a}} \right)$. So the given integral takes the form
$\begin{gathered} \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = \int {\frac{{a\cosh tdt}}{{\sqrt {{a^2}{{\sinh }^2}t + {a^2}} }}} } \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = \int {\frac{{a\cosh tdt}}{{a\sqrt {1 + {{\sinh }^2}t} }}} } \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = \int {\frac{{\cosh tdt}}{{\sqrt {{{\cosh }^2}t} }}} } = \int {dt} \\ \Rightarrow \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = t + c} \\ \end{gathered}$

Using the value $t = {\sin ^{ – 1}}\left( {\frac{x}{a}} \right)$, we have
$\Rightarrow \int {\frac{{dx}}{{\sqrt {{x^2} + {a^2}} }} = {{\sinh }^{ – 1}}\left( {\frac{x}{a}} \right) + c}$