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

Put `v = cos^-1 2x`

(since this is the only substitution that works. The other "likely" one, `v=1-4x^2`, doesn't give us anything useful when we differentiate while doing the integral. I'm using `v` this time, so as not to confuse things with `u` in the following formula.)

We need to find `(dv)/dx`.

In general, (from 3. Derivatives of the Inverse Trigonometric Functions):

`(d(cos^-1u))/(dx)=(-1)/sqrt(1-u^2)(du)/(dx)`

In this example, `u=2x`, so we have `(du)/dx = 2`.

Thus `(dv)/(dx) = (d(cos^-1 2x))/(dx)= (dv)/(du) (du)/dx ` `= (-1)/sqrt(1-(2x)^2)(2) = (-2)/(sqrt[1-4x^2])`

Now our integral doesn't have `-2` as a constant anywhere, but it does have `1/(sqrt[1-4x^2])dx`, so we'll write our differential expression as follows, by dividing throughout by `-2`:

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

So

`int((cos^-1 2x)^4)/(sqrt[1-4x^2]) dx = - 1/2 int v^4 dv`

`=(-1/2) ((v^5)/(5))+K`

`=(-(cos^-1 2x)^5)/(10)+K`

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