Confusing Integral!

• Feb 5th 2011, 04:08 PM
Vamz
Confusing Integral!
$\displaystyle \int \sqrt{4 x - x^2} \, dx$

I really don't know how to evaluate this! I cant get it into the form of any of the inverse trig functions. Im trying to learn the trig substitution method. Could this work on this? Would the trig sub method ever work if the integral was in the form of
$\sqrt{x^2-a^2}$

What method do I use to solve this?
Thank you very much

Heres what I've tried so far

$\int \sqrt{-(x^2-(\sqrt{4x})^2)}dx$
let $x=\sqrt{4x}\tan\theta, dx=\frac{4}{\sqrt{4x}}\sec^2\thetad\theta$

now, I have
$\int \sqrt{-(4x\tan\theta)^2}\frac{4}{\sqrt{4x}}\sec^2\thetad\ theta$

How am I supposed to remove whats in that root? Regardless of what trig function I use, there is always going to be a negative number in it!

Where do I go from here?

Thanks!
• Feb 5th 2011, 04:56 PM
zzzoak
$
x^2-4x+4-4=(x-2)^2-4.
$
• Feb 5th 2011, 05:13 PM
skeeter
$\sqrt{4x - x^2} = \sqrt{4 - (x^2 - 4x + 4)} = \sqrt{4 - (x-2)^2}$

let $x-2 = 2\sin{t}$
• Feb 5th 2011, 07:12 PM
Vamz
Thanks. Can you tell me if I am on the right track here?
so,

$
\displaystyle\int\sqrt{4-(x-2)^2}dx
$

$
(x-2)=2\sin\theta, x=2\sin\theta+2, dx=2\cos\theta d\theta
$

$
\displaystyle \int\sqrt{4-(2\sin\theta)^2} dx
$

$
\displaystyle 2\int\sqrt{1-\sin^2\theta} dx
$

replace dx...
$
\displaystyle 2\int\cos\theta dx = 2\int\cos\theta 2\cos\theta d\theta= 4\int\cos\theta\cos\theta d\theta
$

now, apply half-angle formula
$
\displaystyle 2\int1+\cos(2\theta)d\theta
$

Now, integrate and I get:

$
\displaystyle 2\theta + \sin(2\theta) + C
$

Now, I have to convert theta -> x

$
\displaystyle\theta=\arcsin(\frac{x-2}{2})
$

$
2\arcsin(\frac{x-2}{2}) + \sin(2(\arcsin(\frac{x-2}{2}))) + C
$

could this be correct?
• Feb 5th 2011, 07:46 PM
Prove It
You can clean this up more if you remember that $\displaystyle \sin{2\theta} = 2\sin{\theta}\cos{\theta} = 2\sin{\theta}\sqrt{1 - \sin^2{\theta}}$.

Substitute $\displaystyle \sin{\theta} = \frac{x - 2}{2}$.
• Feb 5th 2011, 09:47 PM
Vamz
I think I am getting the hang of this...

Can you please confirm if this is correct?

$
2\arcsin(\frac{x-2}{2})+(x-2)(\frac{\sqrt{4-(x-2)^2}}{2})
$
• Feb 6th 2011, 10:26 AM
Vamz
Apparently this is not the most simplified form, and there is more than one possibility!
What should I do differently?

Thank you!
• Feb 6th 2011, 11:02 AM
tom@ballooncalculus
Relative simplicity isn't always objective. Are you trying to get an online program to validate your answer? You could use

$\displaystyle{\frac{1}{2}\sqrt{4x - x^2}}$

$\displaystyle{\frac{\sqrt{4 - (x - 2)^2}}{2}}$

or the other slight variation here integrate sqrt&#40;4x - x&#94;2&#41; - Wolfram|Alpha, 'show steps'. (In fact it does some more alarming or at any rate less simple versions at the end.)

Oh, have you included an integration constant? (Wink)
• Feb 6th 2011, 11:10 AM
skeeter
Quote:

Originally Posted by Vamz
I think I am getting the hang of this...

Can you please confirm if this is correct?

$
2\arcsin(\frac{x-2}{2})+(x-2)(\frac{\sqrt{4-(x-2)^2}}{2})
$

+ C and that would be correct.