int_0^1dx/(sqrt(2x-x^2))=? 

প্রশ্ন: \( \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = ? \)

সমাধান:

আমরা প্রথমে ইন্টিগ্র‍্যান্ডটিকে সরল করি:

\( 2x - x^2 = 1 - (1 - 2x + x^2) = 1 - (x - 1)^2 \)

সুতরাং, \( \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = \int_{0}^{1} \frac{dx}{\sqrt{1 - (x - 1)^2}} \)

এখন, ধরি \( x - 1 = \sin(\theta) \). তাহলে, \( dx = \cos(\theta) d\theta \).

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

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

তাহলে, ইন্টিগ্রালটি হবে:

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

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

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

অতএব, \( \int_{0}^{1} \frac{dx}{\sqrt{2x - x^2}} = \frac{\pi}{2} \). 🎉