The key identity is
[tex]\cos^2x=\dfrac{1+\cos2x}2[/tex]
We have
[tex]\cos^4x\sin^2x=\cos^4x(1-\cos^2x)[/tex]
[tex]=\cos^4x-\cos^6x[/tex]
[tex]=\left(\dfrac{1+\cos2x}2\right)^2-\left(\dfrac{1+\cos2x}2\right)^3[/tex]
[tex]=\dfrac{1+\cos2x-\cos^22x-\cos^32x}8[/tex]
Then
[tex]\cos^22x=\dfrac{1+\cos4x}2[/tex]
and
[tex]\cos^32x=\cos2x\cos^22x=\dfrac{\cos2x(1+\cos4x)}2[/tex]
[tex]\cos^32x=\dfrac{\cos2x+\frac{\cos6x+\cos2x}2}2=\dfrac{3\cos2x+\cos6x}4[/tex]
[tex]\cos^4x\sin^2x=\dfrac{1+\cos2x-\frac{1+\cos4x}2-\frac{3\cos2x+\cos6x}4}8[/tex]
[tex]=\dfrac{2+\cos2x-2\cos4x-\cos6x}{32}[/tex]
