int_0^1sqrt((1-x)/(1+x))dx  এর মান কত?

Explanation:

Another Explanation (5): ```html

সমাধান:

ধরি, \(I = \int_0^1 \sqrt{\frac{1-x}{1+x}} dx\) 😊 এখন, \(x = \cos(2\theta)\) ধরলে, \(dx = -2\sin(2\theta) d\theta\) হবে। 🤓 যখন \(x = 0\), \(\cos(2\theta) = 0 \implies 2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}\) 🥳 আবার, যখন \(x = 1\), \(\cos(2\theta) = 1 \implies 2\theta = 0 \implies \theta = 0\) 😎 সুতরাং, \(I = \int_{\frac{\pi}{4}}^0 \sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}} (-2\sin(2\theta)) d\theta\) \(I = \int_0^{\frac{\pi}{4}} \sqrt{\frac{2\sin^2(\theta)}{2\cos^2(\theta)}} (2\sin(2\theta)) d\theta\) \(I = \int_0^{\frac{\pi}{4}} \tan(\theta) \cdot 2 \cdot 2\sin(\theta)\cos(\theta) d\theta\) \(I = \int_0^{\frac{\pi}{4}} \frac{\sin(\theta)}{\cos(\theta)} \cdot 4\sin(\theta)\cos(\theta) d\theta\) \(I = \int_0^{\frac{\pi}{4}} 4\sin^2(\theta) d\theta\) \(I = 4 \int_0^{\frac{\pi}{4}} \frac{1-\cos(2\theta)}{2} d\theta\) \(I = 2 \int_0^{\frac{\pi}{4}} (1-\cos(2\theta)) d\theta\) \(I = 2 \left[ \theta - \frac{\sin(2\theta)}{2} \right]_0^{\frac{\pi}{4}}\) \(I = 2 \left[ \frac{\pi}{4} - \frac{\sin(\frac{\pi}{2})}{2} - (0 - 0) \right]\) \(I = 2 \left[ \frac{\pi}{4} - \frac{1}{2} \right]\) \(I = \frac{\pi}{2} - 1\) 🎉 অতএব, \(\int_0^1 \sqrt{\frac{1-x}{1+x}} dx = \frac{\pi}{2} - 1\) 💖 ```