help with this integral please

[DJ Polaris]

integral of (one over (square root of( -x^2+2x))dx)

 

[stapel]

According to The Integrator, the integral of 1/Sqrt[-x^2+2x] is:


. . . . .\(\displaystyle \large{\frac{2\sqrt{x\,-\,2}\,\log{(\sqrt{x\,-\,2}\,+\,\sqrt{x})}}{\sqrt{2\,-\,x}}}\)


Don't ask me (right now) how to get this, though.... :shock:

Eliz.

 

[DJ Polaris]

Thanks, I kinda need to know how to get there though, but still thanks

 

[stapel]

I kinda need to know how to get there though...

Yeah....

I'll try to work on it, but no promises for tonight: it's kinda late. But at least now you can check any answer you come up with. "The Integrator" can be handy for that. :wink:

Eliz.

 

[Unco]

integral of (one over (square root of( -x^2+2x))dx)

It's a classic trig substitution, if you're familiar.

\(\displaystyle \mbox{ -x^2 + 2x = -(x - 1)^2 + 1 }\)

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

Let \(\displaystyle \mbox{ \sin{\theta} = x - 1}\) if you are unsure.

 

[soroban]

Hello, DJ Polaris!

I think the Integrator is off . . .
How can a real function have both \(\displaystyle \sqrt{x\,-\,2}\) and \(\displaystyle \sqrt{2\,-\,x}\) ?

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

Under the radical, complete the square:

\(\displaystyle \;\;\;1\,-\,x^2\,+\,2x\,-\,1\;=\;1\,-(x^2\,-\,2x\,+\,1)\;=\;1\,-\,(x\,-\,1)^2\)

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

which is of the form: \(\displaystyle \L\int\frac{du}{\sqrt{1\,-\,u^2}}\)\(\displaystyle \;=\;\sin^{-1}(u)\,+\,C\)

 

[stapel]

I think the Integrator is off...

DJ Polaris, I have to agree with the above: I sat down and tried what I was going to recommend (before The Integrator spat out that big ugly thing), and I got the same result that Unco and Soroban got. And differentiating back gives me what we started with.

Sorry 'bout that. The Integrator is generally pretty reliable.

Eliz.

 

[DJ Polaris]

Thanks everyone!!

 

[DJ Polaris]

I think the Integrator is off...

DJ Polaris, I have to agree with the above: I sat down and tried what I was going to recommend (before The Integrator spat out that big ugly thing), and I got the same result that Unco and Soroban got. And differentiating back gives me what we started with.

Sorry 'bout that. The Integrator is generally pretty reliable.

Eliz.

It's ok, no harm done, besides, the Integrator is helpful anyway, so even if that problem was bad, you still helped, so thanks