∫dx/√(a²-x²)=?

আমরা জানি, \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \)। 🤔 এসো, প্রমাণ করি!

ধরি, \( x = a\sin\theta \)। 🤓

তাহলে, \( dx = a\cos\theta d\theta \)। ✍️

এখন,

\( \int \frac{dx}{\sqrt{a^2 - x^2}} = \int \frac{a\cos\theta d\theta}{\sqrt{a^2 - a^2\sin^2\theta}} \)

\(= \int \frac{a\cos\theta d\theta}{\sqrt{a^2(1 - \sin^2\theta)}} \)

\(= \int \frac{a\cos\theta d\theta}{\sqrt{a^2\cos^2\theta}} \)

\(= \int \frac{a\cos\theta d\theta}{a\cos\theta} \)

\(= \int d\theta \)

\(= \theta + C \)

যেহেতু \( x = a\sin\theta \), তাই \( \sin\theta = \frac{x}{a} \)। 😮

সুতরাং, \( \theta = \sin^{-1}\left(\frac{x}{a}\right) \)। 😉

অতএব, \( \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\left(\frac{x}{a}\right) + C \)। 🎉

সুতরাং, উত্তর: \( \sin^{-1}\left(\frac{x}{a}\right) + C \). ✅