intsqrt((1-x)/(1+x)) dx=?

Explanation:

Another Explanation (5): সমাধান: 🤔 ধরি, \(x = \cos{2\theta}\) 🤩 তাহলে, \(dx = -2\sin{2\theta} d\theta\) 😇 এখন, \(\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos{2\theta}}{1+\cos{2\theta}}} = \sqrt{\frac{2\sin^2{\theta}}{2\cos^2{\theta}}} = \sqrt{\tan^2{\theta}} = \tan{\theta}\) 🥳 সুতরাং, \(\int \sqrt{\frac{1-x}{1+x}} dx = \int \tan{\theta} (-2\sin{2\theta}) d\theta\) 😎 \(= -2 \int \tan{\theta} (2\sin{\theta}\cos{\theta}) d\theta = -4 \int \frac{\sin{\theta}}{\cos{\theta}} \sin{\theta} \cos{\theta} d\theta\) 🤓 \(= -4 \int \sin^2{\theta} d\theta = -4 \int \frac{1-\cos{2\theta}}{2} d\theta = -2 \int (1-\cos{2\theta}) d\theta\) 🤯 \(= -2 [\theta - \frac{\sin{2\theta}}{2}] + c = -2\theta + \sin{2\theta} + c\) 🤠 যেহেতু \(x = \cos{2\theta}\), তাহলে \(2\theta = \cos^{-1}{x}\), সুতরাং \(\theta = \frac{1}{2}\cos^{-1}{x}\) 🫡 তাহলে, \(-2\theta + \sin{2\theta} + c = -\cos^{-1}{x} + \sin{(\cos^{-1}{x})} + c\) 🤤 আমরা জানি, \(\sin{(\cos^{-1}{x})} = \sqrt{1-x^2}\). 😎 অতএব, \(-\cos^{-1}{x} + \sqrt{1-x^2} + c\) 🥰 আবার, \(\sin^{-1}{x} + \cos^{-1}{x} = \frac{\pi}{2}\). সুতরাং, \(-\cos^{-1}{x} = \sin^{-1}{x} - \frac{\pi}{2}\) 😈 তাহলে, \(\sin^{-1}{x} - \frac{\pi}{2} + \sqrt{1-x^2} + c = \sin^{-1}{x} + \sqrt{1-x^2} + c'\), যেখানে \(c' = c - \frac{\pi}{2}\) একটি নতুন ধ্রুবক। 😴 সুতরাং, \(\int \sqrt{\frac{1-x}{1+x}} dx = \sin^{-1}{x} + \sqrt{1-x^2} + c\) 🥳