∫ 1 √ x 2 − a 2 dx - Mathematics

Sum
$\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }$

Solution

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

$\text{ Putting x = a sec θ }$

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

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

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

$= \int\sec\tan θ \text{ dθ}$

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

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

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

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

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

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

$\text{ where C' = C }- \text{ ln a }$

Concept: Indefinite Integral Problems
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RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Revision Excercise | Q 42 | Page 203