∫(cosx)^2 *(sinx)^4 *dx
令A=∫(cosx)^2 *(sinx)^4 *dx =∫(cosx)*(sinx)^4*d(sinx)=∫(cosx)*(1/5)*d[(sinx)^5] =(1/5)∫(cosx)*d[(sinx)^5] =(1/5){[cosx*(sinx)^5]-∫(sinx)^5*d(cosx)} =(1/5){[cosx*(sinx)^5]+∫(sinx)^6*dx} =(1/5){[cosx*(sinx)^5]+∫(sinx)^2*(sinx)^4*dx} =(1/5){[cosx*(sinx)^5]+∫(1-(cosx)^2](sinx)^4*dx} =(1/5){[cosx*(sinx)^5]+∫(sinx)^4*dx-∫(cosx)^2(sinx)^4*dx} =(1/5){[cosx*(sinx)^5]+∫(sinx)^4*dx-A} =(1/5){[cosx*(sinx)^5]+∫(sinx)^4*dx}-(A/5) 所以,A=(1/6){[cosx*(sinx)^5]+∫(sinx)^4*dx}…………(1) 而,令B=∫(sinx)^4*dx=-∫(sinx)^3*d(cosx) =-{(sinx)^3*(cosx)-∫(cosx)^2*3*(sinx)^2]dx} =-(sinx)^3*(cosx)+3∫(sinx)^2*[1-(sinx)^2]dx =-(sinx)^3*(cosx)+3∫[(sinx)^2-(sinx)^4]*dx =-(sinx)^3*(cosx)+3∫[1-(cos2x)]/2dx-3∫(sinx)^4*dx =-(sinx)^3*(cosx)+(3x)/2-(3sin2x)/4-3B 所以,B=(-1/4)(sinx)^3*(cosx)+(3x)/8-(3sin2x)/16……(2) 将(2)代入(1)得到: A=(1/6){[cosx*(sinx)^5]+(-1/4)(sinx)^3*(cosx)+(3x)/8-(3sin2x)/16} =(1/6)cosx*(sinx)^5-(1/24)cosx*(sinx)^3+(x/16)-(sin2x)/32。