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

ধরি, \(I = \int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx\)

এখানে, \(\sin^{-1}x = \theta\) ধরলে, \(x = \sin\theta\)

সুতরাং, \(dx = \cos\theta d\theta\)

এবং \(\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta\)

যখন \(x = 0\), \(\theta = \sin^{-1}(0) = 0\)

যখন \(x = 1\), \(\theta = \sin^{-1}(1) = \frac{\pi}{2}\)

অতএব,

\(I = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\cos\theta} \cos\theta d\theta\)

\(= \int_{0}^{\frac{\pi}{2}} \theta d\theta\)

\(= \left[ \frac{\theta^2}{2} \right]_{0}^{\frac{\pi}{2}}\)

\(= \frac{1}{2} \left[ \left(\frac{\pi}{2}\right)^2 - (0)^2 \right]\)

\(= \frac{1}{2} \cdot \frac{\pi^2}{4}\)

\(= \frac{\pi^2}{8}\) 🎉

সুতরাং, \(\int_{0}^{1} \frac{\sin^{-1}x}{\sqrt{1-x^2}} dx = \frac{\pi^2}{8}\) 😊