int_0^(π/2)(1+cosx)²sinxdx = ?

Explanation:

Another Explanation (5): সমাধান: ধরি, \(cosx = z\) 🤓 তাহলে, \(-sinxdx = dz\) অথবা, \(sinxdx = -dz\) সীমা পরিবর্তন করি: যখন \(x = 0\), তখন \(z = cos0 = 1\) যখন \(x = \frac{π}{2}\), তখন \(z = cos\frac{π}{2} = 0\) সুতরাং, \( \int_0^{\frac{π}{2}} (1+cosx)^2 sinxdx = \int_1^0 (1+z)^2 (-dz) \)🤩 \( = -\int_1^0 (1+2z+z^2) dz \) \( = -[z + z^2 + \frac{z^3}{3}]_1^0 \) \( = -[(0+0+0) - (1+1+\frac{1}{3})] \)🥳 \( = -[0 - (2+\frac{1}{3})] \) \( = 2+\frac{1}{3} \) \( = \frac{6+1}{3} \) \( = \frac{7}{3} \)🎉 অতএব, \( \int_0^{\frac{π}{2}} (1+cosx)^2 sinxdx = \frac{7}{3} \) 😎