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

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{1} \frac{\cos^{-1}x}{\sqrt{1-x^2}} dx\) এখানে, আমরা \(u = \cos^{-1}x\) প্রতিস্থাপন করি। তাহলে, \(x = \cos u\) এবং \(dx = -\sin u du\)। যখন \(x = 0\), তখন \(u = \cos^{-1}(0) = \frac{\pi}{2}\)। যখন \(x = 1\), তখন \(u = \cos^{-1}(1) = 0\)। সুতরাং, \(I = \int_{\frac{\pi}{2}}^{0} \frac{u}{\sqrt{1-\cos^2 u}} (-\sin u) du\) \(= \int_{\frac{\pi}{2}}^{0} \frac{u}{\sqrt{\sin^2 u}} (-\sin u) du\) \(= \int_{\frac{\pi}{2}}^{0} \frac{u}{\sin u} (-\sin u) du\) (যেহেতু \(0 \le u \le \frac{\pi}{2}\), তাই \(\sin u \ge 0\)) \(= \int_{\frac{\pi}{2}}^{0} -u du\) \(= -\int_{\frac{\pi}{2}}^{0} u du\) \(= \int_{0}^{\frac{\pi}{2}} u du\) \(= \left[ \frac{u^2}{2} \right]_{0}^{\frac{\pi}{2}}\) \(= \frac{(\frac{\pi}{2})^2}{2} - \frac{0^2}{2}\) \(= \frac{\frac{\pi^2}{4}}{2}\) \(= \frac{\pi^2}{8}\) অতএব, \(\int_{0}^{1} \frac{\cos^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8}\)।🎉