int_0^1 Cos^-1x/sqrt(1-x^2) dx=? 

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_0^1 \frac{\cos^{-1}x}{\sqrt{1-x^2}} dx\) এখানে, \(x = \cos \theta\) ধরলে, \(dx = -\sin \theta d\theta\) হবে। যখন \(x = 0\), \(\theta = \frac{\pi}{2}\) এবং যখন \(x = 1\), \(\theta = 0\). সুতরাং, \(I = \int_{\pi/2}^0 \frac{\cos^{-1}(\cos \theta)}{\sqrt{1-\cos^2 \theta}} (-\sin \theta) d\theta\) \( = \int_{\pi/2}^0 \frac{\theta}{\sqrt{\sin^2 \theta}} (-\sin \theta) d\theta\) \( = \int_{\pi/2}^0 \frac{\theta}{\sin \theta} (-\sin \theta) d\theta\) \( = -\int_{\pi/2}^0 \theta d\theta\) \( = \int_0^{\pi/2} \theta d\theta\) \( = \left[ \frac{\theta^2}{2} \right]_0^{\pi/2}\) \( = \frac{(\pi/2)^2}{2} - \frac{0^2}{2}\) \( = \frac{\pi^2}{8}\) অতএব, \(\int_0^1 \frac{\cos^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8}\) 🥳