求∫(sinx)^4*(cosx)^2dx的积分

Question

如题所述

Answer 1

方法如下，
请作参考：

Answer 2

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

Answer 3

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