1. ## Evaluate this integral

$
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx
$

Can someone please point out where I went wrong?

First, I complete the square:
$
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx
$

Factor out a 16:
$
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx
$

$
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx
$

Let U = $\frac{x-2}{4}$ & dU=1\4dx :: dx=4du

$
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU
$

$
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU
$

Which is:
$
arccosh(U) = arccosh(\frac{x-2}{4})
$

It seems like my answer is wrong however. Where did I go wrong?
Thanks!

2. Originally Posted by Vamz
$
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx
$

Can someone please point out where I went wrong?

First, I complete the square:
$
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx
$

Factor out a 16:
$
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx
$

$
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx
$

Let U = $\frac{x-2}{4}$ & dU=1\4dx :: dx=4du

$
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU
$

$
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU
$

Which is:
$
arccosh(U) = arccosh(\frac{x-2}{4})
$

It seems like my answer is wrong however. Where did I go wrong?
Thanks!
$12+4x-x^2\neq (x-2)^2-16$. However, $12+4x-x^2=16-(x-2)^2$...

3. Originally Posted by Vamz
$
\displaystyle \int \frac{1}{\sqrt{12 + 4 x - x^2}}dx
$

Can someone please point out where I went wrong?

First, I complete the square:
$
\displaystyle \int \frac{1}{\sqrt{(x-2)^2-16}}dx
$

Factor out a 16:
$
\displaystyle \int \frac{1}{\sqrt{16(\frac{(x-2)^2}{16}-1)}}dx
$

$
\displaystyle \frac{1}{4}\int \frac{1}{\sqrt{\frac{(x-2)^2}{16}-1}}dx
$

Let U = $\frac{x-2}{4}$ & dU=1\4dx :: dx=4du

$
\displaystyle \frac{1}{4}\int \frac{4du}{\sqrt{U^2-1}}dU
$

$
\displaystyle \int \frac{du}{\sqrt{U^2-1}}dU
$

Which is:
$
arccosh(U) = arccosh(\frac{x-2}{4})
$

It seems like my answer is wrong however. Where did I go wrong?
Thanks!
First when you completed the square something is off

$12+4x-x^2=-(x^2-4x) +12 =-(x^2-4x+4)+12+4= 16-(x-2)^2$

From here make the substition $x-2=4\sin(t) \implies dx=4\cos(t)dt$

$\displaystyle \int \frac{dx}{\sqrt{12+4x-x^2}}=\int \frac{4\cos(t)dt}{\sqrt{16-16\sin^2(t)}}=\frac{4}{4}\int \frac{\cos(t)dt}{\sqrt{1-\sin^2(t)}}=\int dt$

4. ^ Thank you!

I fixed that problem. Are you using trig substitution to solve this? I havnt learned that very well yet.

Couldnt I still get this into the same form as arccos using the same method I attempted, but this time with an answer of:

arccos((x-2)/4) + C

?

5. $\displaystyle\int{\frac{1}{\sqrt{12+4x-x^2}}}dx=\int{\frac{1}{\sqrt{4^2-\left(x^2-4x+4\right)}}}dx$

$=\displaystyle\int{\frac{1}{\sqrt{4^2-(x-2)^2}}}dx$

$u=x-2\Rightarrow\ du=dx$

$\displaystyle\int{\frac{1}{\sqrt{4^2-u^2}}}du=arcsin\left(\frac{u}{4}\right)+C$

Change the numerator sign of the function to get arccos.

6. Originally Posted by TheEmptySet
First when you completed the square something is off

$12+4x-x^2=-(x^2-4x) +12 =-(x^2-4x+4)+12+4= 16-(x-2)^2$

From here make the substition $x-2=\sin(t) \implies dx=\cos(t)dt$

$\displaystyle \int \frac{dx}{\sqrt{12+4x-x^2}}$

$=\displaystyle\int \frac{4\cos(t)dt}{\sqrt{16-16\sin^2(t)}}$

This doesn't work as the constant under the square root was a 16
rather than a 1

$=\displaystyle\frac{4}{4}\int \frac{\cos(t)dt}{\sqrt{1-\sin^2(t)}}=\int dt$
There's a problem on the 2nd last line.

7. Thanks guys. I understand now

8. the last line is actually correct but I wrote down the wrong sub it should have been

$x-2=4\sin(t) \implies dx=4\cos(t)dt$ which is what I used in the post. :\