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

ধরি, \(x = \sin\theta\), তাহলে \(dx = \cos\theta d\theta\)।

যখন \(x = 0\), \(\theta = 0\); যখন \(x = 1\), \(\theta = \frac{\pi}{2}\)।

সুতরাং, ইন্টিগ্রালটি হবে:

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

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

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

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

\( = \left[ \frac{\theta^3}{3} \right]_0^{\frac{\pi}{2}}\)

\( = \frac{(\frac{\pi}{2})^3}{3} - \frac{0^3}{3}\)

\( = \frac{\pi^3}{8 \times 3}\)

\( = \frac{\pi^3}{24}\)

অতএব, \(\int_0^1 \frac{(\sin^{-1}x)^2}{\sqrt{1-x^2}} dx = \frac{\pi^3}{24}\) 🎉