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प्रश्न
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उत्तर
\[\text{ Let I }= \int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx}\]
\[\text{ Putting x = a sec θ } \]
\[ \Rightarrow \text{ dx = a sec θ tan θ \text{ dθ}} \]
\[ \therefore I = \int\frac{a \sec\theta \tan θ \text{ dθ} }{\sqrt{a^2 \sec^2 \theta - a^2}}\]
\[ = \int\frac{{a \sec\theta\tan θ \text{ dθ} }}{a \cdot \tan\theta}\]
\[ = \int\sec\tan θ \text{ dθ} \]
\[ = \text{ ln }\left| \sec\theta + \tan\theta \right| + C\]
\[ = \text{ ln} \left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\]
\[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\left( \frac{x}{a} \right)^2 - 1} \right| + C\]
\[ = \text{ ln} \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| + C\]
\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \text{ ln a} + C\]
\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| + C'\]
\[\text{ where C' = C }- \text{ ln a }\]
