int(sqrt(1-x)/(1+x))dx  এর মান হচ্ছে- 

Explanation:

Another Explanation (5): সমাধান: ধরি, \( x = \cos{2\theta} \) 🧐 তাহলে, \( dx = -2\sin{2\theta} d\theta \) এখন, \(\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}} \) \(= \sqrt{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}} \) \(= \sqrt{\tan^2{\theta}} = \tan{\theta} \) অতএব, \(\int \sqrt{\frac{1-x}{1+x}} dx = \int \tan{\theta} (-2\sin{2\theta}) d\theta \) \(= -2 \int \tan{\theta} (2\sin{\theta}\cos{\theta}) d\theta \) \(= -4 \int \frac{\sin{\theta}}{\cos{\theta}} \sin{\theta}\cos{\theta} d\theta \) \(= -4 \int \sin^2{\theta} d\theta \) \(= -4 \int \frac{1-\cos{2\theta}}{2} d\theta \) \(= -2 \int (1-\cos{2\theta}) d\theta \) \(= -2 \left[ \theta - \frac{\sin{2\theta}}{2} \right] + c \) \(= -2\theta + \sin{2\theta} + c \) \(= -2\theta + \sqrt{1-\cos^2{2\theta}} + c \) \(= -2\theta + \sqrt{1-x^2} + c \) যেহেতু, \( x = \cos{2\theta} \), তাই \( 2\theta = \cos^{-1}{x} \) সুতরাং, \( \theta = \frac{1}{2} \cos^{-1}{x} \) তাহলে, \( \int \sqrt{\frac{1-x}{1+x}} dx = -\cos^{-1}{x} + \sqrt{1-x^2} + c \) আবার, \( \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \) সুতরাং, \( \int \sqrt{\frac{1-x}{1+x}} dx = -(\frac{\pi}{2} - \sin^{-1}x) + \sqrt{1-x^2} + c \) \( = \sin^{-1}x + \sqrt{1-x^2} + (c-\frac{\pi}{2}) \) \( = \sin^{-1}x + \sqrt{1-x^2} + c' \) [যেখানে \( c' = c - \frac{\pi}{2} \)] অতএব, \( \int \sqrt{\frac{1-x}{1+x}} dx = \sin^{-1}x + \sqrt{1-x^2} + c \) 🎉