int_0^1dx/(sqrt(x-x^2)) 

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

প্রথমে, ইন্টিগ্রালটিকে একটু সরল করা যাক:

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

এখন, \( x = \sin^2(\theta) \) প্রতিস্থাপন করি। তাহলে, \( dx = 2\sin(\theta)\cos(\theta) d\theta \) হবে। যখন \( x = 0 \), \( \theta = 0 \) এবং যখন \( x = 1 \), \( \theta = \frac{\pi}{2} \)। সুতরাং,

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

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

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

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