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, \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \, dx = \frac{2}{3}\). 🥳
