# Math Help - Integral again

1. ## Integral again

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

I start by substituting $x-1=t$ 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 $3-t^2$...I don't see it working...

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

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

Now sub

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

and use the identity

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

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

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

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

or I messed it?

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

Now remember

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

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

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:

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

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

$du = dx$

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

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

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

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

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

Replacing, we get:

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

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

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

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

Done!

After some backsubbing and simplification:

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