int_0^(π/2) cos³x √sinx dx এর মান কোনটি?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(I = \int_{0}^{\frac{\pi}{2}} \cos^3 x \sqrt{\sin x} \, dx\) আমরা জানি, \(\cos^2 x = 1 - \sin^2 x\) সুতরাং, \(I = \int_{0}^{\frac{\pi}{2}} (1 - \sin^2 x) \cos x \sqrt{\sin x} \, dx\) এখন, ধরি \(\sin x = t^2\), সুতরাং \(\cos x \, dx = 2t \, dt\) যখন \(x = 0\), তখন \(t = \sqrt{\sin 0} = 0\) যখন \(x = \frac{\pi}{2}\), তখন \(t = \sqrt{\sin \frac{\pi}{2}} = 1\) অতএব, \(I = \int_{0}^{1} (1 - t^4) \sqrt{t^2} \cdot 2t \, dt\) \(I = 2 \int_{0}^{1} (1 - t^4) t \cdot t \, dt\) \(I = 2 \int_{0}^{1} (t^2 - t^6) \, dt\) \(I = 2 \left[ \frac{t^3}{3} - \frac{t^7}{7} \right]_{0}^{1}\) \(I = 2 \left[ \frac{1}{3} - \frac{1}{7} \right]\) \(I = 2 \left[ \frac{7 - 3}{21} \right]\) \(I = 2 \cdot \frac{4}{21}\) \(I = \frac{8}{21}\) সুতরাং, \(\int_{0}^{\frac{\pi}{2}} \cos^3 x \sqrt{\sin x} \, dx = \frac{8}{21}\) ✅