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

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

The integration of \frac{1}{{x\sqrt {{x^2} - {a^2}} }} is of the form

\int  {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}dx = } \frac{1}{a}{\sec ^{ - 1}}\left(  {\frac{x}{a}} \right) + c


To prove this formula, putting x = a\sec t, we have dx = a\sec t\tan tdt, t = {\sec ^{ - 1}}\left( {\frac{x}{a}} \right), so the given integral takes of the form

\begin{gathered} \int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }}  = \int {\frac{{a\sec t\tan tdt}}{{a\sec t\sqrt {{a^2}{{\sec }^2}t - {a^2}} }}}  } \\ \Rightarrow \int {\frac{{dx}}{{x\sqrt {{x^2}  - {a^2}} }} = \int {\frac{{\sin tdt}}{{\sqrt {{a^2}{{\sec }^2}t - {a^2}} }}}  } \\ \Rightarrow \int {\frac{{dx}}{{x\sqrt {{x^2}  - {a^2}} }} = \int {\frac{{\tan tdt}}{{a\sqrt {{{\sec }^2}t} }} =  \frac{1}{a}\int {\frac{{\tan tdt}}{{\tan t}}} } }  = \int {dt} \\ \Rightarrow \int {\frac{{dx}}{{x\sqrt {{x^2}  - {a^2}} }} = \frac{1}{a}t + c} \\ \end{gathered}


Using the value t =  {\sec ^{ - 1}}\left( {\frac{x}{a}} \right), we have

  \Rightarrow \int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }} = \frac{1}{a}{{\sec  }^{ - 1}}\left( {\frac{x}{a}} \right) + c}

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