int_0^(pi/2)cos x /sqrt(4-sin^2x) dx = কত?

প্রশ্ন: \(\int_0^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{4 - \sin^2 x}} \, dx = ?\)

সমাধান:

ধরি, \(u = \sin x\), তাহলে \(du = \cos x \, dx\)।

যখন \(x = 0\), \(u = \sin 0 = 0\).

যখন \(x = \frac{\pi}{2}\), \(u = \sin \frac{\pi}{2} = 1\).

সুতরাং, ইন্টিগ্রালটি হবে:

\(\int_0^1 \frac{1}{\sqrt{4 - u^2}} \, du\)

আমরা জানি, \(\int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + C\).

এখানে, \(a = 2\).

সুতরাং, \(\int_0^1 \frac{1}{\sqrt{4 - u^2}} \, du = \left[ \sin^{-1} \left( \frac{u}{2} \right) \right]_0^1\)

\(= \sin^{-1} \left( \frac{1}{2} \right) - \sin^{-1} \left( \frac{0}{2} \right)\)

\(= \sin^{-1} \left( \frac{1}{2} \right) - \sin^{-1} (0)\)

আমরা জানি, \(\sin^{-1} \left( \frac{1}{2} \right) = \frac{\pi}{6}\) এবং \(\sin^{-1} (0) = 0\).

সুতরাং, \(\frac{\pi}{6} - 0 = \frac{\pi}{6}\).

অতএব, \(\int_0^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{4 - \sin^2 x}} \, dx = \frac{\pi}{6}\).

উত্তর: \(\frac{\pi}{6}\) 🎉