Question

大神帮我把这个积分求下！

Answer 1

追问

错了吧？

4次方呢？

追答

好像漏了

∫(sinx)^4dx=∫(sinx)^2*(sinx)^2dx=∫((1/2)*(1-cos2x))*((1/2)*(1-cos2x))dx
=∫(1/4)*(1+(cos2x)^2-2cos2x)dx=(1/4)x+(1/4)∫(cos2x)^2dx-(1/4)sin2x
=(1/4)x+(1/8)∫(cos4x+1)dx-(1/4)sin2x
∫(sinx)^4dx=∫(sinx)^2*(sinx)^2dx=∫((1/2)*(1-cos2x))*((1/2)*(1-cos2x))dx
=∫(1/4)*(1+(cos2x)^2-2cos2x)dx=(1/4)x+(1/4)∫(cos2x)^2dx-(1/4)sin2x
=(1/4)x+(1/8)∫(cos4x+1)dx-(1/4)sin2x
=(3/8)x+(1/32)sin4x-(1/4)sin2x+c