Integral by completing the square

[jared944]

Ok, I know this problem shouldn't be driving me insane, but it is.


Im looking for the indefinate integral of 1/(2x-x^2)^(1/2)

I cant seem to get the denominator that I need. Here's what I have so far -

Under the radical, I have

2x-x^2
Pull out the -1

-1 (x^2+2x)
a) .5 * 2 = 1
b) 1^2 = 1

So,

-1 (x^2 +2x +1)
-1 (x + 1)^2 - 1

Does that work thusfar?

Thanks,

Jared[/img]

 

[mark07]

-1 (x^2 - 2x)
-1 (x^2 - 2x + 1 - 1)
-1 (x - 1)^2 + 1

= 1 - (x - 1)^2

Then you have

\(\displaystyle \L \int \frac{1}{\sqrt{1 - (x - 1)^2}} \, dx\)

Now use the trig substitution,

\(\displaystyle \L x-1 = \sin \theta\)

Can you finish after this?

 

[jared944]

outstanding, thank you!

 

[galactus]

Let \(\displaystyle \L\\u=x-1, \;\ du=dx, \;\ u+1=x\)

The substitutions lead to:

\(\displaystyle \L\\\int\frac{1}{\sqrt{1-u^{2}}}du=sin^{-1}(u)\)

Resub the u:

\(\displaystyle \L\\sin^{-1}(x-1)\)