I'm having trouble integrating this: dx/sqrt(5+3x-2x^2). I rearranged it to form dx/sqrt(31/8 - 2(x-3/4)^2) Not sure what to do after that. Should I have not factored the 2 out? Thanks.
You have $\displaystyle -2x^2 + 3x +5 = -2(x^2-\frac{3}{2}x-\frac{5}{2})$
$\displaystyle -2(x^2-\frac{3}{2}x+\frac{9}{16}-\frac{9}{16}-\frac{5}{2})$
$\displaystyle -2((x-\frac{3}{4})^2-\frac{49}{16}) = \frac{49}{8} - 2(x-\frac{3}{4})^2 $
You can now use a trigo substitution to find the answer.
Keeping the 2 in front would be a nice idea.
This leads to $\displaystyle *\int \frac{1}{\sqrt{2}}\frac{dx}{\sqrt{\frac{49}{16}-(x-\frac{3}{4})^2}} $*
You should use $\displaystyle \frac{49}{16}sin(\theta) = x $*
You know these integrals are very tricky and I may make some mistakes.
I'm not too bad at understanding but making no mistakes is not one of my aptitudes.
You can check yourself with the online integrator of mathematica.
It gives $\displaystyle *\frac{1}{\sqrt{2}}arcsin(\frac{4}{7}x-\frac{3}{7}) +C$