int x^2/(√(1-x^2) dx = ? 

Explanation:

Another Explanation (5):

প্রশ্ন: \(\int \frac{x^2}{\sqrt{1-x^2}} dx = ?\)

সমাধান:

ধরি, \(x = \sin(\theta)\), তাহলে \(dx = \cos(\theta) d\theta\) সুতরাং, \(\int \frac{x^2}{\sqrt{1-x^2}} dx = \int \frac{\sin^2(\theta)}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) d\theta\) \(= \int \frac{\sin^2(\theta)}{\sqrt{\cos^2(\theta)}} \cos(\theta) d\theta\) \(= \int \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) d\theta\) \(= \int \sin^2(\theta) d\theta\) আমরা জানি, \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\) সুতরাং, \(\int \sin^2(\theta) d\theta = \int \frac{1 - \cos(2\theta)}{2} d\theta\) \(= \frac{1}{2} \int (1 - \cos(2\theta)) d\theta\) \(= \frac{1}{2} \left[ \int 1 d\theta - \int \cos(2\theta) d\theta \right]\) \(= \frac{1}{2} \left[ \theta - \frac{\sin(2\theta)}{2} \right] + C\) \(= \frac{1}{2} \theta - \frac{1}{4} \sin(2\theta) + C\) আমরা জানি, \(\sin(2\theta) = 2\sin(\theta)\cos(\theta)\) সুতরাং, \(= \frac{1}{2} \theta - \frac{1}{4} (2\sin(\theta)\cos(\theta)) + C\) \(= \frac{1}{2} \theta - \frac{1}{2} \sin(\theta)\cos(\theta) + C\) যেহেতু \(x = \sin(\theta)\), তাই \(\theta = \arcsin(x)\) এবং \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)} = \sqrt{1 - x^2}\) অতএব, \(= \frac{1}{2} \arcsin(x) - \frac{1}{2} x \sqrt{1 - x^2} + C\) সুতরাং, \(\int \frac{x^2}{\sqrt{1-x^2}} dx = \frac{1}{2} \arcsin(x) - \frac{1}{2} x \sqrt{1 - x^2} + C\) 🎉