∫_0^(pi/2)(1+cosx)2sinxdx এর মান কত?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(cosx = z\) সুতরাং, \(-sinxdx = dz\) \(sinxdx = -dz\) সীমা পরিবর্তন করে পাই, যখন \(x = 0\), \(z = cos0 = 1\) যখন \(x = \frac{\pi}{2}\), \(z = cos\frac{\pi}{2} = 0\) অতএব, \(\int_0^{\frac{\pi}{2}} (1+cosx)^2 sinxdx = \int_1^0 (1+z)^2 (-dz)\) \( = -\int_1^0 (1+2z+z^2) dz\) \( = -[z + z^2 + \frac{z^3}{3}]_1^0\) \( = -[(0+0+0) - (1+1+\frac{1}{3})]\) \( = -[0 - (2+\frac{1}{3})]\) \( = 2 + \frac{1}{3}\) \( = \frac{6+1}{3}\) \( = \frac{7}{3}\) সুতরাং, \(\int_0^{\frac{\pi}{2}} (1+cosx)^2 sinxdx = \frac{7}{3}\) 🥳