求定积分∫(sinx)^(n-1)cos(n+1)xdx,上限为π,下限为0.书上说用分部积分法

∫[0,π] sinx^(n-1) cosx^(n+1)dx=∫[0,π]sinx^(n-1)cosx^(n-1)*cosx^2dx=(1/2^n)∫[0,π](sin2x)^n [(1+cos2x)/2 ]dx= (1/2^n)∫[0,π]sin(2x)^ndx - (1/2^(n+2))∫[0,π]sin(2x)^ndsin2x=(1/2^(n+1))∫[0,π]sin...