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`
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