int_0^(1/2) (dx)/(sqrt(1-x^2))  এর মান কোনটি ?  

আমরা জানি, \( \int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}(x) + C \)।

সুতরাং, \( \int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}} = \left[ \sin^{-1}(x) \right]_0^{\frac{1}{2}} \) 🤓।

এখন, \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \) এবং \( \sin^{-1}(0) = 0 \)।

অতএব, \( \int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}\left(\frac{1}{2}\right) - \sin^{-1}(0) = \frac{\pi}{6} - 0 = \frac{\pi}{6} \) 🎉।

সুতরাং, \( \int_0^{\frac{1}{2}} \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{6} \) 😎।