Question

\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx}\]

\[\text{  Putting  x = a  sec θ }  \]

\[ \Rightarrow \text{ dx = a sec θ   tan   θ   \text{  dθ}} \]

\[ \therefore I = \int\frac{a \sec\theta \tan  θ    \text{ dθ} }{\sqrt{a^2 \sec^2 \theta - a^2}}\]

\[ = \int\frac{{a \sec\theta\tan  θ    \text{ dθ} }}{a \cdot \tan\theta}\]

\[ = \int\sec\tan  θ    \text{ dθ} \]

\[ = \text{ ln }\left| \sec\theta + \tan\theta \right| + C\]

\[ = \text{ ln} \left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\]

\[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\left( \frac{x}{a} \right)^2 - 1} \right| + C\]

\[ = \text{ ln} \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| + C\]

\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \text{ ln a} + C\]

\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| + C'\]

\[\text{ where C'  = C }- \text{ ln  a }\]