int_0^(π/2)cos³ x dx = ? 

Explanation:

Another Explanation (5):

সমাধান: \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx\)

আমরা জানি, \(\int_{0}^{\frac{\pi}{2}} \cos^n x \, dx = \frac{n-1}{n} \cdot \frac{n-3}{n-2} \cdots \frac{2}{3} \cdot 1\) যদি \(n\) বিজোড় সংখ্যা হয়। 🎉 এখানে, \(n = 3\), সুতরাং: \[ \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = \frac{3-1}{3} = \frac{2}{3} \] অন্যভাবে, আমরা \(\cos^3 x\) কে \(\cos x\) এর মাধ্যমে প্রকাশ করতে পারি: 🤓 \[ \cos^3 x = \cos x \cdot \cos^2 x = \cos x (1 - \sin^2 x) \] তাহলে, \[ \int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = \int_{0}^{\frac{\pi}{2}} \cos x (1 - \sin^2 x) \, dx \] 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, \[ \int_{0}^{\frac{\pi}{2}} \cos x (1 - \sin^2 x) \, dx = \int_{0}^{1} (1 - u^2) \, du \] \[ = \left[ u - \frac{u^3}{3} \right]_{0}^{1} = \left( 1 - \frac{1}{3} \right) - (0 - 0) = 1 - \frac{1}{3} = \frac{2}{3} \] অতএব, \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = \frac{2}{3}\). 🥳