int_0^((√3)/2)frac((sin^-1x)^2)sqrt(1-x^2)dx=? 

প্রশ্ন: \(\int_0^{\frac{\sqrt{3}}{2}} \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}} dx = ?\)

উত্তর: \(\frac{\pi^3}{81}\)

সমাধান:

1. ধরি, \(x = \sin \theta\)। সুতরাং, \(dx = \cos \theta d\theta\)। 😊
2. যখন \(x = 0\), \(\theta = \sin^{-1}(0) = 0\)।
3. যখন \(x = \frac{\sqrt{3}}{2}\), \(\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}\)।
4. তাহলে, ইন্টিগ্রালটি হবে: \[ \int_0^{\frac{\pi}{3}} \frac{(\sin^{-1}(\sin \theta))^2}{\sqrt{1-\sin^2 \theta}} \cos \theta d\theta = \int_0^{\frac{\pi}{3}} \frac{\theta^2}{\sqrt{\cos^2 \theta}} \cos \theta d\theta \]
5. যেহেতু \(\cos \theta > 0\) যখন \(\theta \in \left[0, \frac{\pi}{3}\right]\), \(\sqrt{\cos^2 \theta} = \cos \theta\)।
6. সুতরাং, \[ \int_0^{\frac{\pi}{3}} \frac{\theta^2}{\cos \theta} \cos \theta d\theta = \int_0^{\frac{\pi}{3}} \theta^2 d\theta \]
7. এখন, \(\int \theta^2 d\theta = \frac{\theta^3}{3} + C\)।
8. সুতরাং, \[ \int_0^{\frac{\pi}{3}} \theta^2 d\theta = \left[\frac{\theta^3}{3}\right]_0^{\frac{\pi}{3}} = \frac{\left(\frac{\pi}{3}\right)^3}{3} - \frac{0^3}{3} = \frac{\pi^3}{3 \cdot 27} = \frac{\pi^3}{81} \]

অতএব, \(\int_0^{\frac{\sqrt{3}}{2}} \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}} dx = \frac{\pi^3}{81}\)। 🎉