int_0^(pi/2) Cos^3xSinxdx=?

প্রশ্ন: \( \int_{0}^{\frac{\pi}{2}} \cos^3(x) \sin(x) \, dx = ? \) 🤔

সমাধান:

ধরি, \( u = \cos(x) \). 🤓

তাহলে, \( \frac{du}{dx} = -\sin(x) \), সুতরাং \( du = -\sin(x) \, dx \) অথবা \( -du = \sin(x) \, dx \). 🧐

এখন, যখন \( x = 0 \), তখন \( u = \cos(0) = 1 \). 😉

আবার, যখন \( x = \frac{\pi}{2} \), তখন \( u = \cos(\frac{\pi}{2}) = 0 \). 🤩

অতএব, আমাদের ইন্টিগ্রালটি হবে:

\( \int_{0}^{\frac{\pi}{2}} \cos^3(x) \sin(x) \, dx = \int_{1}^{0} u^3 (-du) \) 🤗

\(= -\int_{1}^{0} u^3 \, du = \int_{0}^{1} u^3 \, du \) 👍

\(= \left[ \frac{u^4}{4} \right]_{0}^{1} \) 😎

\(= \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4} - 0 = \frac{1}{4} \)🥳

সুতরাং, \( \int_{0}^{\frac{\pi}{2}} \cos^3(x) \sin(x) \, dx = \frac{1}{4} \). 🎉

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