int_0^(pi/2)(1+cosx)^2sinxdx এর মান-

Explanation:

Another Explanation (5): bài toán tích phân \(\int_0^{\frac{\pi}{2}} (1+\cos x)^2 \sin x \, dx\) ধরি, \(u = 1 + \cos x\). তাহলে, \(\frac{du}{dx} = -\sin x\), সুতরাং \(du = -\sin x \, dx\) বা \(\sin x \, dx = -du\). যখন \(x = 0\), \(u = 1 + \cos 0 = 1 + 1 = 2\). যখন \(x = \frac{\pi}{2}\), \(u = 1 + \cos \frac{\pi}{2} = 1 + 0 = 1\). সুতরাং, ইন্টিগ্রালটি হবে: \(\int_2^1 u^2 (-du) = -\int_2^1 u^2 \, du = \int_1^2 u^2 \, du\) এখন, \(u^2\) এর ইন্টিগ্রাল হলো \(\frac{u^3}{3}\). সুতরাং, \(\int_1^2 u^2 \, du = \left[ \frac{u^3}{3} \right]_1^2 = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\) সুতরাং, \(\int_0^{\frac{\pi}{2}} (1+\cos x)^2 \sin x \, dx = \frac{7}{3}\). উত্তর: \(\frac{7}{3}\) 🎉