int_0^(π/2) cos ^3xsin^2xdx = ? 

Explanation:

Another Explanation (5): আ integral টি সমাধান করা হল: ধরি, \(I = \int_0^{\frac{\pi}{2}} \cos^3(x) \sin^2(x) \, dx\) আমরা \(\cos^3(x)\)-কে \(\cos^2(x) \cdot \cos(x)\) আকারে লিখতে পারি। সুতরাং, \(I = \int_0^{\frac{\pi}{2}} \cos^2(x) \sin^2(x) \cos(x) \, dx\) আমরা জানি, \(\cos^2(x) = 1 - \sin^2(x)\). সুতরাং, \(I = \int_0^{\frac{\pi}{2}} (1 - \sin^2(x)) \sin^2(x) \cos(x) \, dx\) এখন, ধরি \(u = \sin(x)\). তাহলে, \(\frac{du}{dx} = \cos(x)\), অর্থাৎ, \(du = \cos(x) \, dx\). সীমা পরিবর্তন করি: যখন \(x = 0\), \(u = \sin(0) = 0\). যখন \(x = \frac{\pi}{2}\), \(u = \sin(\frac{\pi}{2}) = 1\). সুতরাং, integral টি হবে: \(I = \int_0^1 (1 - u^2) u^2 \, du\) \(I = \int_0^1 (u^2 - u^4) \, du\) এখন integrate করি: \(I = \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1\) \(I = \left( \frac{1^3}{3} - \frac{1^5}{5} \right) - \left( \frac{0^3}{3} - \frac{0^5}{5} \right)\) \(I = \frac{1}{3} - \frac{1}{5}\) \(I = \frac{5 - 3}{15}\) \(I = \frac{2}{15}\) অতএব, \(\int_0^{\frac{\pi}{2}} \cos^3(x) \sin^2(x) \, dx = \frac{2}{15}\) 🎉.