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

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