You should use half angle formulas such that:

`sin^2 x = (1 - cos 2x)/2 => sin^4 x = (1 - cos 2x)^2/4`

`cos^2 x = (1 + cos 2x)/2 => cos^4 x = (1 + cos 2x)^2/4`

`sin^4 x*cos^4 x = (1 - cos 2x)^2/4*(1 + cos 2x)^2/4`

`sin^4 x*cos^4 x = (1 - cos^2 2x)^2/16`

`sin^4 x*cos^4 x = (1 - 2cos^2 2x + cos^4 2x)^2/16`

Integrating both sides yields:

`int sin^4 x*cos^4 x dx= int (1 - 2cos^2 2x + cos^4 2x)^2/16 dx`

Using the linearity property of integrals yields:

`int sin^4 x*cos^4 x dx = (1/16)(int dx - 2int cos^2 2x dx + int cos^4 2x dx)`

`int sin^4 x*cos^4 x dx = (1/16)(int dx - 2int(1+cos 4x)/2 dx + int (1+cos 4x)^2/4 dx)`

`int sin^4 x*cos^4 x dx = (1/16)(int dx - int dx - int cos 4x + (1/4) int (1 + 2cos 4x + cos^2 4x) dx)`

`int sin^4 x*cos^4 x dx = (1/16)(-(sin4x)/4 + (1/4)x + (1/2)(sin4x)/4 + (1/4)int (1 + cos 8x)/2 dx)`

`int sin^4 x*cos^4 x dx = (1/16)(-(sin4x)/4 + (1/4)x + (1/8)x + (sin8x)/64) + c`

`int sin^4 x*cos^4 x dx = (1/16)(-(sin4x)/4+ (3/8)x + (sin8x)/64) + c`

**Hence, evaluating the given indefinite integral yields `int sin^4 x*cos^4 x dx = (1/16)(-(sin4x)/4+ (3/8)x + (sin8x)/64) + c.` **

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