Question

高数微积分，请问这道题怎么做？

Answer 1

∵(sinx)^4(cosx)^2
=(1-cos2x)^2(1+cos2x)/8
=[1-(cos2x)^2](1-cos2x)/8
=(sin2x)^2(1-cos2x)/8
=[1-(cos4x)]/16-(sin2x)^2(cos2x)/8
∴原积分=∫[1-(cos4x)]/16*dx-∫(sin2x)^2(cos2x)/8*dx
=x/16-(sin4x)]/64-1/16*∫(sin2x)^2(dsin2x)
=x/16-(sin4x)]/64-(sin2x)^3/48+C

Answer 2

∫(sinx)^4(cosx)^2dx = (1/8)∫[2(sinx)^2][4(sinx)^2(cosx)^2]dx
= (1/8)∫(1-cos2x)(sin2x)^2dx = (1/8)∫(sin2x)^2dx - (1/8)∫cos2x(sin2x)^2dx
= (1/16)∫(1-cos4x)dx - (1/16)∫(sin2x)^2dsin2x
= x/16 - (1/64)sin4x - (1/48)(sin2x)^3 + C

Answer 3

不断用二倍角公式即可