int_0^1 1/(sqrt(2x-x^2) dx=? 

সমাধান:

আমরা প্রথমে ইন্টিগ্রালটিকে এভাবে লিখতে পারি:

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

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 হবে:

\(\int_{-\pi/2}^0 \frac{\cos\theta}{\sqrt{1 - \sin^2\theta}} d\theta = \int_{-\pi/2}^0 \frac{\cos\theta}{\sqrt{\cos^2\theta}} d\theta = \int_{-\pi/2}^0 \frac{\cos\theta}{\cos\theta} d\theta\)

\(= \int_{-\pi/2}^0 1 d\theta = [\theta]_{-\pi/2}^0 = 0 - (-\frac{\pi}{2}) = \frac{\pi}{2}\)

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