Put `u = 1 - 2x^4`

Then `du = -8x^3dx`

The question has `x^3dx` so we write

`-(du)/8=x^3dx`

So we have:

`int_-1^0x^3(1-2x^4)^3dx=-1/8int_(x=-1)^(x=0)u^3du`

`=-1/8xx[u^4/4]_(x=-1)^(x=0)`

`=-1/32[u^4]_(x=-1)^(x=0)`

`=-1/32[(1-2x^4)^4]_-1^0`

`=-1/32[(1-0)^4-(1-2)^4]`

`=-1/32[(1)-(1)]`

`=0`

Using the 2nd approach

If we take the second approach, i.e. leaving the variable changed, we have:

`u = 1 - 2x^4`

As `x = -1 → 0` (this means "as `x` takes values from `-1` to `0`"),

then `u = -1 → 1`

So the problem becomes:

`int_-1^0x^3(1-2x^4)^3dx =-1/8int_-1^1u^3du`

`=-1/32[u^4]_-1^1`

`=-1/32[1^4-(-1)^4]`

`=0` as before.

This second approach is quite useful later when the substitutions become more involved (e.g. trigonometric substitution).