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प्रश्न
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उत्तर
\[\text{ Let I } = \int\frac{dx}{\sqrt{x^2 - a^2}}\]
\[\text{ Putting x} = a \tan \theta\]
\[ \Rightarrow dx = a \sec^2 \text{ θ dθ }\]
\[ \therefore I = \int\frac{a \cdot se c^2\text{ θ dθ }}{\sqrt{a^2 \tan^2 \theta + a^2}}\]
\[ = \int\frac{a \sec^2 \theta \cdot d\theta}{a\sqrt{1 + \tan^2 \theta}}\]
\[ = \int\frac{\sec^2 \theta \cdot \text{ dθ }}{\sec\theta}\]
\[ = \int\sec\theta \cdot d\theta\]
\[ = \int\sec\theta \cdot d\theta\]
\[ = \text{ ln } \left| \sec\theta + \tan\theta \right| + C\]
\[ = \text{ ln }\left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\]
\[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1} \right| + C\]
\[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \ln a + C\]
\[ = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C'\]
\[\text{ where C' = C - ln a }\]
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संबंधित प्रश्न
Evaluate the following integral:
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
