1. ## Integral again

Need help with this integral!
$\displaystyle \int{\frac{dx}{[(x-1)^2+3]\sqrt{3-(x-1)^2}}}$

I start by substituting $\displaystyle x-1=t$ but don't know how to go further...

2. try u = 3- t
I like dealing with easier roots
Might help here too

3. Originally Posted by jbpellerin
try u = 3- t
I like dealing with easier roots
Might help here too
But here we have $\displaystyle 3-t^2$...I don't see it working...

4. $\displaystyle \int{\frac{dt}{[t^2+3]\sqrt{3-t^2}}}$

$\displaystyle \frac{1}{\sqrt{3}}\int{\frac{dt}{[t^2+3]\sqrt{1-(\frac{t}{\sqrt{3}})^2}}}$

Now sub

$\displaystyle \frac{t}{\sqrt3} =sin\theta$

and use the identity

$\displaystyle 1- sin^2\theta = cos^2\theta$

$\displaystyle \frac{1}{\sqrt{3}}\int \frac{\sqrt{3}cos\theta(d\theta)}{[3sin^2\theta +3]cos\theta} = \int \frac{cos\theta(d\theta)}{[3sin^2\theta +3]cos\theta} = \int\frac{cos\theta(d\theta)}{(3-3cos^2\theta+3)cos\theta}$

$\displaystyle = \int\frac{cos\theta(d\theta)}{(6 -3cos^2\theta)cos\theta}= \int\frac{d\theta}{6-3cos^2\theta} = \int \frac{d\theta}{3(sin^2\theta + 1)}$

5. but by dividing with $\displaystyle \sqrt{3}$ wouldn't it be
$\displaystyle \frac{1}{\sqrt{3}} \int{\frac{dt}{[t^2+3]\sqrt{1-(\frac{t}{\color{red}{\sqrt{3}}}})^2}}$
or I messed it?

6. $\displaystyle \int\frac{\frac{dt}{\sqrt{3}}}{\frac{[t^2+3]\sqrt{3-t^2}}{\sqrt{3}}}$

Now remember

$\displaystyle \frac{\sqrt{3-t^2}}{\sqrt{3}} = \sqrt{\frac{3-t^2}{3}} = \sqrt{1-\bigg(\frac{t}{\sqrt{3}}\bigg)^2}$

7. yes man
$\displaystyle \sqrt{\frac{3-t^2}{3}} = \sqrt{1-\frac{t^2}{3}}$ and not $\displaystyle \sqrt{1-\left(\frac{t}{3}\right)^2}$

8. Thanks for pointing that out. I can't believe I did that, I'm ashamed lol. Ok let me go fix my mistake.

9. Here's my approach:

$\displaystyle \int{\frac{dx}{[(x-1)^2+3]\sqrt{3-(x-1)^2}}}$

First substitution to make things look neat:
$\displaystyle u = x-1$

$\displaystyle du = dx$

$\displaystyle \int{\frac{du}{[u^2+3]\sqrt{3-u^2}}}$

Second substitution:
$\displaystyle t = \frac{1}{u^2+3}$

$\displaystyle dt = -\frac{2u}{(u^2+3)^2}$

$\displaystyle 3-u^2 = \frac{6t-1}{t}$

$\displaystyle u = \sqrt{\frac{1-3t}{t}}$

Replacing, we get:

$\displaystyle -\frac{1}{2} \int \frac{1}{t \sqrt{\frac{1-3t}{t}} \sqrt{\frac{6t-1}{t}}}$

$\displaystyle = -\frac{1}{2} \int \frac{1}{(\sqrt{1-3t})(\sqrt{6t-1})}$

$\displaystyle = -\frac{1}{2} \int \frac{1}{\sqrt{-18t^2+9t-1}}$

By completing the square, we get:
$\displaystyle = -\frac{\sqrt{8}}{2} \int \frac{1}{\sqrt{1-(12t-3)^2}}$

Done!

After some backsubbing and simplification:

$\displaystyle = -\frac{\sqrt{2}}{12} \arcsin{\left(\frac{-3x^2+6x}{(x-1)^2+3}\right)} + C$