int_0^1(2(sin^-1x)^3)/(sqrt(1-x^2)) dx এর মান কত?

ধরি, \(x = \sin(\theta)\). তাহলে, \(dx = \cos(\theta) d\theta\). যখন \(x = 0\), \(\theta = 0\) এবং যখন \(x = 1\), \(\theta = \frac{\pi}{2}\).

সুতরাং, ইন্টিগ্রালটি হবে:

\(\int_0^1 \frac{2(\sin^{-1}x)^3}{\sqrt{1-x^2}} dx = \int_0^{\frac{\pi}{2}} \frac{2(\sin^{-1}(\sin\theta))^3}{\sqrt{1-\sin^2\theta}} \cos\theta d\theta\)

\(= \int_0^{\frac{\pi}{2}} \frac{2\theta^3}{\sqrt{\cos^2\theta}} \cos\theta d\theta = \int_0^{\frac{\pi}{2}} \frac{2\theta^3}{\cos\theta} \cos\theta d\theta\)

\(= \int_0^{\frac{\pi}{2}} 2\theta^3 d\theta = 2 \int_0^{\frac{\pi}{2}} \theta^3 d\theta\)

\(= 2 \left[ \frac{\theta^4}{4} \right]_0^{\frac{\pi}{2}} = 2 \left( \frac{(\frac{\pi}{2})^4}{4} - 0 \right)\)

\(= 2 \cdot \frac{\pi^4}{16 \cdot 4} = \frac{2\pi^4}{64} = \frac{\pi^4}{32}\)

অতএব, \(\int_0^1 \frac{2(\sin^{-1}x)^3}{\sqrt{1-x^2}} dx = \frac{\pi^4}{32}\) 🎉.