integration

How he made this !

$$\int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx
=\int
\frac{1}{\sqrt{4[ 1- \left( \frac{x-3}{2} \right)^2} ] }\ dx
=\frac{1}{2}\int
\frac{1}{\sqrt{1- \left( \frac{x-3}{2} \right)^2}}\ dx\\\\
\small{\text{We substitute:
$
\frac{x-3}{2}=\sin{(u)},
$
so we have
$
\frac{1}{2}\ dx=\cos{(u)}\ du
$
or
$
\ dx=2\cos{(u)}\ du
$
}}\\
=\frac{1}{2}\int
\frac{2\cos{(u)}}{\sqrt{1- \sin^2{(u)}}}\ du
=\frac{1}{2}\int
\frac{2\cos{(u)}}{\cos{(u)}}}\ du
=\int\ du=u\\
\small{\text{We substitute back:
$
u=\arcsin{ ( \frac{x-3}{2} ) }
$
}}\\\\
\int {\frac{1}{\sqrt{4-(x-3)^2}} }\ dx =\sin^{-1}{ ( \frac{x-3}{2} ) }$$

-5 + 6x - x^2 =

-5 - (x^2 - 6x )          complete the square on x...take one-half the coefficient on x

(= 3), square it (= 9)....then......add and subtract it...so we have...

-5 - (x^2 - 6x + 9 - 9) =

-5 + 9 - (x^2 - 6x + 9)       factor  the expression on the parenthesis

4 - (x - 3)^2

And that's it...

But where is my mistake here 

Let's go from this step

-(x^2 -6x + 9 - 9) - 5    =

-(x^2 - 6x + 9) +  9 - 5  =

-(x - 3)^2 + 4  =

4 - (x - 3)^2

But why we cant multiply the negative sign inside the parentheses in this step 

-(x^2 - 6x + 9) +  9 - 5

so its become 

(-x^2 + 6x - 9) + 9 - 5

You mistake is that

$$(x-3)^2=x^2-6x+9\\\\
$It does not equal $-x^2+6x-9\\\
$that is why -1 had to be left out of the bracket!$$$

Thanks, Melody.....

My pleasure Chris :)

Melody 

I don't get it 

$$-x^2+6x-9$$ = $$(x-3)^2$$

Also 

$$x^2-6x+9 = (x-3)^2$$

so what is wrong !!! 

NO

$$-x^2+6x-9$$  does not equal      $$(x-3)^2$$      

$$\\-x^2+6x-9= -(x^2-6x+9)=-1*(x^2-6x+9) = -1*(x-3)^2\\\\
$the -1 is not squared so$ \\\\
(x-3)^2 \ne -x^2+6x-9$$

remember 

    $$-3^2\ne(-3)^2$$

consider these two sums

$$20+-3^2 = 20-3^2=20-9=11$$

and

$$20+ (-3)^2 = 20++9=29$$

see        $$-3^2\ne(-3)^2$$

I hope that helps :)

Now its clear .. thank you