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

Explanation:

Another Explanation (5): সমাধান: 🤔 আমরা প্রথমে ইন্টিগ্রালটিকে একটু সরল করি: \[ \int_0^1 \frac{dx}{\sqrt{2x - x^2}} = \int_0^1 \frac{dx}{\sqrt{1 - (x-1)^2}} \] এখন, \(x - 1 = \sin\theta\) ধরি। তাহলে, \(dx = \cos\theta d\theta\). যখন \(x = 0\), \(\sin\theta = -1\), সুতরাং \(\theta = -\frac{\pi}{2}\). যখন \(x = 1\), \(\sin\theta = 0\), সুতরাং \(\theta = 0\). তাহলে, ইন্টিগ্রালটি হবে: \[ \int_{-\pi/2}^0 \frac{\cos\theta d\theta}{\sqrt{1 - \sin^2\theta}} = \int_{-\pi/2}^0 \frac{\cos\theta d\theta}{\sqrt{\cos^2\theta}} = \int_{-\pi/2}^0 d\theta \] যেহেতু \(-\frac{\pi}{2} \le \theta \le 0\), \(\cos\theta \ge 0\), তাই \(\sqrt{\cos^2\theta} = \cos\theta\). \[ \int_{-\pi/2}^0 d\theta = [\theta]_{-\pi/2}^0 = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} \] সুতরাং, \[ \int_0^1 \frac{dx}{\sqrt{2x - x^2}} = \frac{\pi}{2} \] অতএব, উত্তর \(\frac{\pi}{2}\). 🎉