int dx/(xsqrt(x^2-1)) =?

প্রশ্ন: \(\int \frac{dx}{x\sqrt{x^2-1}} = ?\)

সমাধান:

ধরি, \(x = \sec(\theta)\)
তাহলে, \(dx = \sec(\theta)\tan(\theta) d\theta\)

সুতরাং, \(\int \frac{dx}{x\sqrt{x^2-1}} = \int \frac{\sec(\theta)\tan(\theta) d\theta}{\sec(\theta)\sqrt{\sec^2(\theta)-1}}\)

\(= \int \frac{\sec(\theta)\tan(\theta) d\theta}{\sec(\theta)\sqrt{\tan^2(\theta)}}\)

\(= \int \frac{\sec(\theta)\tan(\theta) d\theta}{\sec(\theta)\tan(\theta)}\)

\(= \int d\theta\)

\(= \theta + c\), যেখানে c একটি ধ্রুবক।

যেহেতু \(x = \sec(\theta)\), তাই \(\theta = \sec^{-1}(x)\)

অতএব, \(\int \frac{dx}{x\sqrt{x^2-1}} = \sec^{-1}(x) + c\)

উত্তর: \(\sec^{-1}x+c\) 🎉