This is the original definite integral. $$\int_1^3\frac{2x-3}{\sqrt{4x-x^2}}$$ I do not know what to do after completing the square and I am stuck at that specific part after plugging back in. $$(4x-x^2)$$ $$-(x^2-4x )$$ $$-(x^2-4x+4)-4$$ $$-(x-2)^2-4$$ So I would plug that in back to the integral $$\int_1^3\frac{2x-3}{\sqrt{-(x-2)^2-4}}$$ and this is where I am not sure what to do next I was thinking u-substitution but what would be my u?
$\endgroup$ 31 Answer
$\begingroup$Oops, I think you made a small algebra slip (no worries, it happens to the best of us!). You need to swap the -4 for +4 in your working since:
$-(x^2-4x) = -(x^2-4x)-4+4 = -(x^2-4x+4)+4 = 4-(x-2)^2$
As for a hint for the rest of it, you could notice that:
$\frac{d} {dx} (4-(x-2)^2) = 4-2x$
Using this you could split you integral into 2 parts by writing the numerator of your fraction as: $ -(4-2x) + 1 $
This would leave two integrals one which you can find an anti derivative for and another which you can use a trig substitution to solve. (Try thinking about $cos^2x + sin^2x = 1$ if you're stuck!) Hope this helps.
