∫(1,0)√(1-x)/(1+x)dx=?  

ধাপ ১: প্রতিস্থাপন ব্যবহার করা যাক। ধরি, \(x = \cos(2\theta)\)। তাহলে, \(dx = -2\sin(2\theta) d\theta\)।

এখন, \(\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} = \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} = \tan(\theta)\)

সীমা পরিবর্তন করি: যখন \(x = 0\), \(\cos(2\theta) = 0\), সুতরাং \(2\theta = \frac{\pi}{2}\) এবং \(\theta = \frac{\pi}{4}\)। যখন \(x = 1\), \(\cos(2\theta) = 1\), সুতরাং \(2\theta = 0\) এবং \(\theta = 0\)।

সুতরাং, ইন্টিগ্রালটি হবে:

\(\int_{\frac{\pi}{4}}^{0} \tan(\theta) \cdot (-2\sin(2\theta)) d\theta = 2 \int_{0}^{\frac{\pi}{4}} \tan(\theta) \cdot 2\sin(\theta)\cos(\theta) d\theta\)

\(= 4 \int_{0}^{\frac{\pi}{4}} \frac{\sin(\theta)}{\cos(\theta)} \cdot \sin(\theta)\cos(\theta) d\theta = 4 \int_{0}^{\frac{\pi}{4}} \sin^2(\theta) d\theta\)

ধাপ ২: \(\sin^2(\theta)\) এর ইন্টিগ্রেশন

\(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\)

সুতরাং, \(4 \int_{0}^{\frac{\pi}{4}} \sin^2(\theta) d\theta = 4 \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2\theta)}{2} d\theta = 2 \int_{0}^{\frac{\pi}{4}} (1 - \cos(2\theta)) d\theta\)

\(= 2 \left[ \theta - \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{4}} = 2 \left[ \left( \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]\)

\(= 2 \left[ \frac{\pi}{4} - \frac{1}{2} \right] = \frac{\pi}{2} - 1\)

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