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...
$\displaystyle \int{\frac{dt}{[t^2+3]\sqrt{3-t^2}}}
$
Divide by radical 3
$\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} $
$\displaystyle = \int\frac{cos\theta(d\theta)}{(6 -3cos^2\theta)cos\theta}= \int\frac{d\theta}{6-3cos^2\theta} = \int \frac{d\theta}{3(sin^2\theta + 1)}$
$\displaystyle \int\frac{\frac{dt}{\sqrt{3}}}{\frac{[t^2+3]\sqrt{3-t^2}}{\sqrt{3}}}$
Now remember
$\displaystyle \frac{\sqrt{3-t^2}}{\sqrt{3}} = \sqrt{\frac{3-t^2}{3}} = \sqrt{1-\bigg(\frac{t}{\sqrt{3}}\bigg)^2}$
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:
$\displaystyle = -\frac{\sqrt{2}}{12} \arcsin{\left(\frac{-3x^2+6x}{(x-1)^2+3}\right)} + C$