int_0^(π/2) (cosxdx)/(sqrt(4-sin^2x))=?

প্রশ্ন: \( \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{4 - \sin^2 x}} dx = ? \)

সমাধান:

ধরি, \( \sin x = u \)

তাহলে, \( \cos x \, dx = du \)

যখন \( x = 0 \), তখন \( u = \sin 0 = 0 \)

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

সুতরাং, সমাকলনটি হবে:

\( \int_{0}^{1} \frac{du}{\sqrt{4 - u^2}} \)

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

এখানে, \( a = 2 \)

সুতরাং, \( \int_{0}^{1} \frac{du}{\sqrt{4 - u^2}} = \left[ \sin^{-1} \left( \frac{u}{2} \right) \right]_{0}^{1} \)

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

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

\(= \frac{\pi}{6} - 0 \)

\(= \frac{\pi}{6} \)

অতএব, \( \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sqrt{4 - \sin^2 x}} dx = \frac{\pi}{6} \)

উত্তর: \(\frac{\pi}{6}\) 🎉