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?

  • $\begingroup$ First, need to add dx's to your integrals. Second, have you thought about making a trig substitution? $\endgroup$ – David Sep 20 '15 at 21:48
  • $\begingroup$ Hint First, note that you have a mistake since $4x-x^2=4-(x-2)^2$. Second, use that $$ \int \frac{du}{\sqrt{1-u^2}}=\sin^{-1}(u)+C,$$ in your case you will have to split the integral and use $u=(x-2)/2$ $\endgroup$ – Alonso Delfín Sep 20 '15 at 22:02
  • $\begingroup$ I have another question lets say i was taking the integral of another equation which required me to complete the square and these were the values $x^4+x^2+C$ when i factor out to complete the square would I also factor out a $x^2$? leading it to this $x^2(x^2+1)+C$ ? $\endgroup$ – Carlos V Sep 21 '15 at 1:07

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.

  • $\begingroup$ Would you mind elaborating on why it would be +4 instead of -4 on the outside i am a bit confused on that. $\endgroup$ – Carlos V Sep 21 '15 at 0:57
  • $\begingroup$ Oh never mind, another user pointed it out thank you anyways. $\endgroup$ – Carlos V Sep 21 '15 at 1:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.