int_0^1 (sin^-1x)/(sqrt(1-x^2))dx 

প্রশ্ন: \(\int_0^1 \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx\)

উত্তর: \(\frac{\pi^2}{8}\)

সমাধান:

ধরি, \(I = \int_0^1 \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx\)

এখানে, \(x = \sin\theta\) প্রতিস্থাপন করি। তাহলে, \(dx = \cos\theta d\theta\) হবে।

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

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

সুতরাং, \(I = \int_0^{\frac{\pi}{2}} \frac{\theta}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta\)

\(\implies I = \int_0^{\frac{\pi}{2}} \frac{\theta}{\sqrt{\cos^2\theta}} \cos\theta d\theta\)

\(\implies I = \int_0^{\frac{\pi}{2}} \frac{\theta}{\cos\theta} \cos\theta d\theta\)

\(\implies I = \int_0^{\frac{\pi}{2}} \theta d\theta\)

এখন, সমাকলন করে পাই,

\(I = \left[\frac{\theta^2}{2}\right]_0^{\frac{\pi}{2}}\)

\(I = \frac{(\frac{\pi}{2})^2}{2} - \frac{0^2}{2}\)

\(I = \frac{\pi^2}{4 \times 2}\)

\(I = \frac{\pi^2}{8}\)

অতএব, \(\int_0^1 \frac{\sin^{-1}x}{\sqrt{1-x^2}}dx = \frac{\pi^2}{8}\) 🎉🎉🎉