int_0^1 dx/(sqrt(4-3x²))=? 

Explanation:

Another Explanation (5): সমাধান: 🤔 $$\int_{0}^{1} \frac{dx}{\sqrt{4-3x^2}}$$ ধরি, \(x = \frac{2}{\sqrt{3}} \sin{\theta}\) 🤓 তাহলে, \(dx = \frac{2}{\sqrt{3}} \cos{\theta} d\theta\) 🧐 যখন \(x = 0\), \(\sin{\theta} = 0\), সুতরাং \(\theta = 0\) 😇 যখন \(x = 1\), \(\sin{\theta} = \frac{\sqrt{3}}{2}\), সুতরাং \(\theta = \frac{\pi}{3}\) 🤩 অতএব, $$\int_{0}^{1} \frac{dx}{\sqrt{4-3x^2}} = \int_{0}^{\pi/3} \frac{\frac{2}{\sqrt{3}} \cos{\theta} d\theta}{\sqrt{4 - 3(\frac{4}{3} \sin^2{\theta})}}$$ $$= \int_{0}^{\pi/3} \frac{\frac{2}{\sqrt{3}} \cos{\theta} d\theta}{\sqrt{4 - 4\sin^2{\theta}}}$$ $$= \int_{0}^{\pi/3} \frac{\frac{2}{\sqrt{3}} \cos{\theta} d\theta}{\sqrt{4(1 - \sin^2{\theta})}}$$ $$= \int_{0}^{\pi/3} \frac{\frac{2}{\sqrt{3}} \cos{\theta} d\theta}{2\cos{\theta}}$$ $$= \frac{1}{\sqrt{3}} \int_{0}^{\pi/3} d\theta$$ $$= \frac{1}{\sqrt{3}} [\theta]_{0}^{\pi/3}$$ $$= \frac{1}{\sqrt{3}} \left(\frac{\pi}{3} - 0\right)$$ $$= \frac{\pi}{3\sqrt{3}}$$ সুতরাং, \(\int_{0}^{1} \frac{dx}{\sqrt{4-3x^2}} = \frac{\pi}{3\sqrt{3}}\) 🎉