Evaluate : int_0^(π/2) cos^3 xsqrt(sinx) dx   

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_0^{\frac{\pi}{2}} \cos^3 x \sqrt{\sin x} \, dx\) আমরা \(\cos^3 x = \cos^2 x \cdot \cos x = (1 - \sin^2 x) \cos x\) লিখতে পারি। তাহলে, \(I = \int_0^{\frac{\pi}{2}} (1 - \sin^2 x) \sqrt{\sin x} \cos x \, dx\) এখন, ধরি \(u = \sin x\)। তাহলে, \(du = \cos x \, dx\)। যখন \(x = 0\), তখন \(u = \sin 0 = 0\). যখন \(x = \frac{\pi}{2}\), তখন \(u = \sin \frac{\pi}{2} = 1\). সুতরাং, \(I = \int_0^1 (1 - u^2) \sqrt{u} \, du = \int_0^1 (1 - u^2) u^{\frac{1}{2}} \, du\) \( = \int_0^1 (u^{\frac{1}{2}} - u^{\frac{5}{2}}) \, du\) \( = \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} - \frac{u^{\frac{7}{2}}}{\frac{7}{2}} \right]_0^1\) \( = \left[ \frac{2}{3} u^{\frac{3}{2}} - \frac{2}{7} u^{\frac{7}{2}} \right]_0^1\) \( = \left( \frac{2}{3} (1)^{\frac{3}{2}} - \frac{2}{7} (1)^{\frac{7}{2}} \right) - \left( \frac{2}{3} (0)^{\frac{3}{2}} - \frac{2}{7} (0)^{\frac{7}{2}} \right)\) \( = \frac{2}{3} - \frac{2}{7} = \frac{14 - 6}{21} = \frac{8}{21}\) অতএব, \(\int_0^{\frac{\pi}{2}} \cos^3 x \sqrt{\sin x} \, dx = \frac{8}{21}\). 🎉