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

Explanation:

Another Explanation (5): সমাধান: ধরি, \( I = \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx \) 🤔 এখন, \( \sin^{-1}x = \theta \) ধরলে, \( x = \sin\theta \) হবে। সুতরাং, \( dx = \cos\theta d\theta \) 😊 সীমা পরিবর্তন করে পাই, যখন \( x = 0 \), তখন \( \theta = \sin^{-1}(0) = 0 \) এবং যখন \( x = 1 \), তখন \( \theta = \sin^{-1}(1) = \frac{\pi}{2} \) 🤓 তাহলে, \( I = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta \) 😎 \( = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sqrt{\cos^2\theta}} \cos\theta d\theta \) \( = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\cos\theta} \cos\theta d\theta \) \( = \int_{0}^{\frac{\pi}{2}} \theta d\theta \) 🤩 \( = \left[ \frac{\theta^2}{2} \right]_{0}^{\frac{\pi}{2}} \) \( = \frac{1}{2} \left[ \left(\frac{\pi}{2}\right)^2 - 0^2 \right] \) \( = \frac{1}{2} \cdot \frac{\pi^2}{4} \) \( = \frac{\pi^2}{8} \) 🎉 অতএব, \( \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8} \) 🥰