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

Explanation:

Another Explanation (5):

প্রশ্ন: \(\int_0^1 \frac{dx}{\sqrt{2x-x^2}} = ?\)

সমাধান:

আমরা প্রথমে ইন্টিগ্রালটিকে এভাবে লিখতে পারি:

\[ \int_0^1 \frac{dx}{\sqrt{2x-x^2}} = \int_0^1 \frac{dx}{\sqrt{1 - (x^2 - 2x + 1)}} = \int_0^1 \frac{dx}{\sqrt{1 - (x-1)^2}} \]

এখন, \(x-1 = \sin\theta\) ধরলে, \(dx = \cos\theta d\theta\) হবে।

যখন \(x = 0\), \(\sin\theta = -1\), সুতরাং \(\theta = -\frac{\pi}{2}\).

যখন \(x = 1\), \(\sin\theta = 0\), সুতরাং \(\theta = 0\).

তাহলে ইন্টিগ্রালটি দাঁড়ায়:

\[ \int_{-\pi/2}^0 \frac{\cos\theta d\theta}{\sqrt{1 - \sin^2\theta}} = \int_{-\pi/2}^0 \frac{\cos\theta d\theta}{\sqrt{\cos^2\theta}} = \int_{-\pi/2}^0 \frac{\cos\theta d\theta}{|\cos\theta|} \]

যেহেতু \(-\frac{\pi}{2} \le \theta \le 0\) এর মধ্যে \(\cos\theta\) ধনাত্মক, তাই \(|\cos\theta| = \cos\theta\).

সুতরাং,

\[ \int_{-\pi/2}^0 \frac{\cos\theta}{\cos\theta} d\theta = \int_{-\pi/2}^0 d\theta = [\theta]_{-\pi/2}^0 = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2} \]

অতএব, \(\int_0^1 \frac{dx}{\sqrt{2x-x^2}} = \frac{\pi}{2}\) 🥳.