int_0^(pi/2)(1+cosx)^2sinxdx

 এর মান নির্ণয় করঃ 

ধরি, \(I = \int_0^{\frac{\pi}{2}} (1+\cos x)^2 \sin x \, dx\)

এখন, ধরি \(1+\cos x = z\)

তাহলে, \(\frac{dz}{dx} = -\sin x\)

সুতরাং, \(dz = -\sin x \, dx\)

সুতরাং, \(\sin x \, dx = -dz\)

যখন \(x = 0\), তখন \(z = 1 + \cos 0 = 1 + 1 = 2\)

যখন \(x = \frac{\pi}{2}\), তখন \(z = 1 + \cos \frac{\pi}{2} = 1 + 0 = 1\)

তাহলে, \(I = \int_2^1 z^2 (-dz)\)

\(= -\int_2^1 z^2 dz\)

\(= \int_1^2 z^2 dz\)

\(= \left[\frac{z^3}{3}\right]_1^2\)

\(= \frac{2^3}{3} - \frac{1^3}{3}\)

\(= \frac{8}{3} - \frac{1}{3}\)

\(= \frac{7}{3}\) 🎉

অতএব, \(\int_0^{\frac{\pi}{2}} (1+\cos x)^2 \sin x \, dx = \frac{7}{3}\)