`int 2 / sqrt(-x^2+4x) dx` Find or evaluate the integral by completing the square

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marizi eNotes educator| Certified Educator

To evaluate the given integral:` int 2/sqrt(-x^2+4x)dx` , we  may apply the basic integration property: `int c*f(x)dx= c int f(x)dx` .

The integral becomes:

`2 int dx/sqrt(-x^2+4x)`

We complete the square for the expression `(-x^2+4x)` .

Completing the square:

For the first step, factor out (-1): `(-x^2+4x) = (-1)(x^2-4x) or -(x^2-4x)`

The `x^2 -4x ` or `x^-4x+0` resembles the `ax^2+bx+c ` where:

`a=1` , `b =-4` and `c=0` .

To complete the square, we add and subtract `(-b/(2a))^2` .

Using `a=1` and `b=-4` , we get:

`(-b/(2a))^2 =(-(-4)/(2(1)))^2`

              `=(4/2)^2`

              ` = 2^2`

              `=4`

Add and subtract 4 inside the` (x^2-4x)` :

`-(x^2-4x+4 -4)`

Distribute the negative sign on -4 to rewrite it as:

`-(x^2-4x+4) +4`

Factor the perfect square trinomial: `x^2-4x+4 = (x-2)^2` .

`-(x-2)^2 +4`

 

For the original problem, we let: `-x^2+4x=-(x-2)^2 +4` :

`2 int dx/sqrt(-x^2+4x)=2 int dx/sqrt(-(x-2)^2+4)`

It can also be rewritten as:

`2 int dx/sqrt(-(x-2)^2 +2^2) =2 int dx/sqrt(2^2 -(x-2)^2)`

The integral part resembles the integral formula:

`int (du)/sqrt(a^2-u^2) = arcsin(u/a)+C` .

Applying the formula, we get:

`2 int dx/sqrt(2^2 -(x-2)^2) =2 *(arcsin (x-2)/2) +C`

 Then the indefinite integral :

`int 2/sqrt(-x^2+4x)dx = 2arcsin((x-2)/2)+C`