int_0^1 (sin^-1x)/(sqrt(1-x^2))dx এর মান কত?

ধরি, \(I = \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx\)

এখানে, আমরা \( \sin^{-1}x = \theta \) প্রতিস্থাপন করি।

তাহলে, \( x = \sin\theta \) এবং \( dx = \cos\theta d\theta \) হবে।

সীমা পরিবর্তন করে পাই, যখন \( x = 0 \), তখন \( \theta = \sin^{-1}(0) = 0 \) এবং যখন \( x = 1 \), তখন \( \theta = \sin^{-1}(1) = \frac{\pi}{2} \) হবে।

সুতরাং, \( I = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta \)

\( = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sqrt{\cos^2\theta}} \cos\theta d\theta \)

\( = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\cos\theta} \cos\theta d\theta \)

\( = \int_{0}^{\frac{\pi}{2}} \theta d\theta \)

\( = \left[ \frac{\theta^2}{2} \right]_{0}^{\frac{\pi}{2}} \)

\( = \frac{(\frac{\pi}{2})^2}{2} - \frac{0^2}{2} \)

\( = \frac{\frac{\pi^2}{4}}{2} \)

\( = \frac{\pi^2}{8} \)

অতএব, \( \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8} \) 🥳