`int (sin^-1 4x)/(sqrt[1-16x^2]) dx`

We have some choices for u in this example. Either `sin^-1 4x`, or `1 − 16x^2`, or `sqrt(1 − 16x^2)`. Only one of these gives a result for du that we can use to integrate the given expression, and that's the first one.

So we let `u=sin^-1 4x`

Then, using the derivative of the inverse sine, we have:

`du=(4)/(sqrt[1-16x^2]) dx`

We divide both sides by 4 so we can substitue into our original expression:

`1/4 du=(1)/(sqrt[1-16x^2]) dx`

Now to complete the required subsitution (u = sin^-14x and the `(du)/4` expression we just found):

`int(sin^-1 4x)/(sqrt[1-16x^2]) dx=1/4intu\ du`

The expression on the right is a simple integral:

`1/4intu\ du = 1/4((u^2)/(2))+K`

To complete the problem, we substitute sin^-14x for u:

`int(sin^-1 4x)/(sqrt[1-16x^2]) dx = ((sin^-1 4x)^2)/(8)+K`