∫_1^1 (dx)/(4-x^2)^(1/2) 

প্রশ্ন: \(\int_1^2 \frac{dx}{\sqrt{4-x^2}}\)

সমাধান:

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

সুতরাং, \(\int_1^2 \frac{dx}{\sqrt{4-x^2}} = \int_1^2 \frac{dx}{\sqrt{2^2-x^2}} = \left[ \sin^{-1}\left(\frac{x}{2}\right) \right]_1^2 \)

= \(\sin^{-1}\left(\frac{2}{2}\right) - \sin^{-1}\left(\frac{1}{2}\right)\)

= \(\sin^{-1}(1) - \sin^{-1}\left(\frac{1}{2}\right)\)

আমরা জানি, \(\sin^{-1}(1) = \frac{\pi}{2}\) এবং \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\)

অতএব, \(\frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi - \pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}\)

সুতরাং, \(\int_1^2 \frac{dx}{\sqrt{4-x^2}} = \frac{\pi}{3}\) 🎉🎉🎉

উত্তর: \(\frac{\pi}{3}\)