∫(0,π)√[sinx-(sinx)^3]dx =∫(0,π)√[sinx(cosx)^2] =∫(0,π/2)cosx√sinxdx-∫(π/2,π)cosx√sinxdx =∫(0,π/2)√sinxdsinx-∫(π/2,π)√sinxdsinx =(2/3)(sinx)^(3/2)-(2/3)(sinx)^(3/2) =(2/3)(sinπ/2)^(3/2)-(2/3)(sin0)^(3/2)-(2/3)(sinπ)^(3/2)+(2/3)(sinπ/2)^(3/2) =2/3-0-0+2/3 =4/3