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

Explanation:

Another Explanation (5): সমাধান: \[ \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} \] আমরা \(2x - x^2\) কে লিখতে পারি: \[ 2x - x^2 = 1 - (1 - 2x + x^2) = 1 - (x - 1)^2 \] সুতরাং, সমাকলটি হবে: \[ \int_{0}^{1} \frac{dx}{\sqrt{1 - (x - 1)^2}} \] এখন, ধরি \(x - 1 = \sin\theta\). তাহলে, \(dx = \cos\theta d\theta\). যখন \(x = 0\), \(\sin\theta = -1\), সুতরাং \(\theta = -\frac{\pi}{2}\). যখন \(x = 1\), \(\sin\theta = 0\), সুতরাং \(\theta = 0\). অতএব, সমাকলটি হবে: \[ \int_{-\pi/2}^{0} \frac{\cos\theta d\theta}{\sqrt{1 - \sin^2\theta}} = \int_{-\pi/2}^{0} \frac{\cos\theta d\theta}{\sqrt{\cos^2\theta}} = \int_{-\pi/2}^{0} \frac{\cos\theta d\theta}{\cos\theta} = \int_{-\pi/2}^{0} d\theta \] \[ = [\theta]_{-\pi/2}^{0} = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} \] সুতরাং, \[ \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = \frac{\pi}{2} \] 🎉সুতরাং নির্ণেয় মান \(\frac{\pi}{2}\)।🥳