int_0^1(sin^-1x)/(sqrt(1-x^2)dx=?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_0^1 \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx\) এখন, \( \sin^{-1}x = \theta \) ধরি। সুতরাং, \( x = \sin\theta \) তাহলে, \( dx = \cos\theta d\theta \) যখন \( x = 0 \), তখন \( \theta = \sin^{-1}(0) = 0 \) যখন \( x = 1 \), তখন \( \theta = \sin^{-1}(1) = \frac{\pi}{2} \) সুতরাং, সমাকলনটি হবে: \( I = \int_0^{\frac{\pi}{2}} \frac{\theta}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta \) \( = \int_0^{\frac{\pi}{2}} \frac{\theta}{\sqrt{\cos^2\theta}} \cos\theta d\theta \) \( = \int_0^{\frac{\pi}{2}} \frac{\theta}{\cos\theta} \cos\theta d\theta \) \( = \int_0^{\frac{\pi}{2}} \theta d\theta \) এখন, \(\theta\) এর সাপেক্ষে সমাকলন করে পাই: \( I = \left[ \frac{\theta^2}{2} \right]_0^{\frac{\pi}{2}} \) \( = \frac{1}{2} \left[ \left(\frac{\pi}{2}\right)^2 - (0)^2 \right] \) \( = \frac{1}{2} \cdot \frac{\pi^2}{4} \) \( = \frac{\pi^2}{8} \) অতএব, \( \int_0^1 \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8} \) 🥳🎉