\(\int_{0}^{1} \left( \sin^{-1} \right)^2 \sqrt{1-x^2} dx\) এর মান কত?

ধরি, \(x = \sin{\theta}\). সুতরাং, \(dx = \cos{\theta} d\theta\).

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

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

তাহলে, ইন্টিগ্রালটি হবে:

\(\int_{0}^{1} (\sin^{-1} x)^2 \sqrt{1-x^2} dx = \int_{0}^{\frac{\pi}{2}} (\sin^{-1} (\sin{\theta}))^2 \sqrt{1-\sin^2{\theta}} \cos{\theta} d\theta\)

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

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

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

আমরা জানি, \(\cos^2{\theta} = \frac{1 + \cos{2\theta}}{2}\).

সুতরাং, ইন্টিগ্রালটি হবে:

\(\int_{0}^{\frac{\pi}{2}} \theta^2 \left(\frac{1 + \cos{2\theta}}{2}\right) d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 \cos{2\theta} d\theta\)

প্রথম ইন্টিগ্রালটি হলো:

\(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 d\theta = \frac{1}{2} \left[ \frac{\theta^3}{3} \right]_{0}^{\frac{\pi}{2}} = \frac{1}{2} \cdot \frac{(\frac{\pi}{2})^3}{3} = \frac{\pi^3}{48}\)

দ্বিতীয় ইন্টিগ্রালটির জন্য, আমরা ইন্টিগ্রেশন বাই পার্টস ব্যবহার করি:

\(\int \theta^2 \cos{2\theta} d\theta = \theta^2 \cdot \frac{\sin{2\theta}}{2} - \int 2\theta \cdot \frac{\sin{2\theta}}{2} d\theta\)

\(= \frac{\theta^2 \sin{2\theta}}{2} - \int \theta \sin{2\theta} d\theta\)

\(= \frac{\theta^2 \sin{2\theta}}{2} - \left[ \theta \left( -\frac{\cos{2\theta}}{2} \right) - \int 1 \cdot \left( -\frac{\cos{2\theta}}{2} \right) d\theta \right]\)

\(= \frac{\theta^2 \sin{2\theta}}{2} + \frac{\theta \cos{2\theta}}{2} - \frac{1}{2} \int \cos{2\theta} d\theta\)

\(= \frac{\theta^2 \sin{2\theta}}{2} + \frac{\theta \cos{2\theta}}{2} - \frac{1}{2} \cdot \frac{\sin{2\theta}}{2}\)

\(= \frac{\theta^2 \sin{2\theta}}{2} + \frac{\theta \cos{2\theta}}{2} - \frac{\sin{2\theta}}{4}\)

সুতরাং,

\(\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 \cos{2\theta} d\theta = \frac{1}{2} \left[ \frac{\theta^2 \sin{2\theta}}{2} + \frac{\theta \cos{2\theta}}{2} - \frac{\sin{2\theta}}{4} \right]_{0}^{\frac{\pi}{2}}\)

\(= \frac{1}{2} \left[ \left( \frac{(\frac{\pi}{2})^2 \sin{\pi}}{2} + \frac{\frac{\pi}{2} \cos{\pi}}{2} - \frac{\sin{\pi}}{4} \right) - \left( 0 + 0 - 0 \right) \right]\)

\(= \frac{1}{2} \left[ 0 + \frac{\pi}{2} \cdot \frac{-1}{2} - 0 \right] = \frac{1}{2} \left[ -\frac{\pi}{4} \right] = -\frac{\pi}{8}\)

তাহলে, সম্পূর্ণ ইন্টিগ্রালটি হলো:

\(\frac{\pi^3}{48} - \frac{\pi}{8} = \frac{\pi^3 - 6\pi}{48}\)

কিন্তু আমাদের দেওয়া উত্তর \(\frac{\pi}{3}\). তাহলে কোথাও ভুল হয়েছে।

আবার চেষ্টা করি।

\(\int_{0}^{\frac{\pi}{2}} \theta^2 \cos^2{\theta} d\theta = \int_{0}^{\frac{\pi}{2}} \theta^2 (\frac{1+\cos 2\theta}{2}) d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 d\theta + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^2 \cos 2\theta d\theta \)

\(= \frac{1}{2}[\frac{\theta^3}{3}]_{0}^{\frac{\pi}{2}} + \frac{1}{2}[\frac{\theta^2}{2} \sin 2\theta + \frac{\theta}{2}\cos 2\theta - \frac{1}{4} \sin 2\theta]_{0}^{\frac{\pi}{2}}\)

\(= \frac{1}{2} \frac{\pi^3}{24} + \frac{1}{2} [0 + \frac{\pi}{4}(-1) - 0 - 0] = \frac{\pi^3}{48} - \frac{\pi}{8} = \frac{\pi^3-6\pi}{48}\)

আবার ইন্টিগ্রেশন বাই পার্টস করি:

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

\(= \int_{0}^{\frac{\pi}{2}} \theta^2 \frac{1+\cos 2\theta}{2} d\theta\)

\(= \frac{1}{2} [\frac{\theta^3}{3}]_{0}^{\frac{\pi}{2}} + \frac{1}{2}\int_{0}^{\frac{\pi}{2}} \theta^2 \cos 2\theta d\theta\)

\(= \frac{\pi^3}{48} + \frac{1}{2}[\theta^2 \frac{\sin 2\theta}{2} - \int 2\theta \frac{\sin 2\theta}{2} d\theta]_{0}^{\frac{\pi}{2}}\)

\(= \frac{\pi^3}{48} + \frac{1}{2}[0 - [\theta (-\frac{\cos 2\theta}{2}) - \int (-\frac{\cos 2\theta}{2})d\theta]_{0}^{\frac{\pi}{2}}\)

\(= \frac{\pi^3}{48} + \frac{1}{2}[\frac{\theta \cos 2\theta}{2} - \frac{\sin 2\theta}{4}]_{0}^{\frac{\pi}{2}}\)

\(= \frac{\pi^3}{48} + \frac{1}{2}[\frac{\pi}{2}\frac{-1}{2} - 0] = \frac{\pi^3}{48} - \frac{\pi}{8}\)

মনে করি, \(x=\cos\theta\) সুতরাং, \(dx = -\sin\theta d\theta\)

যখন \(x = 0\), \(\theta = \frac{\pi}{2}\).

যখন \(x = 1\), \(\theta = 0\).

\(\int_{0}^{1} (\sin^{-1} x)^2 \sqrt{1-x^2} dx = \int_{\frac{\pi}{2}}^{0} (\sin^{-1} (\cos\theta))^2 \sqrt{1-\cos^2\theta} (-\sin\theta) d\theta\)

\(= \int_{0}^{\frac{\pi}{2}} (\sin^{-1} (\cos\theta))^2 \sin^2\theta d\theta\)

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

যদি উত্তর \(\frac{\pi}{3}\) হয়, তবে প্রশ্ন অথবা উত্তরের কোথাও ভুল আছে।