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 October 28th, 2010, 03:17 PM #1 Newbie   Joined: Oct 2010 Posts: 3 Thanks: 0 Impossible to integrate? I am taking Calculus II right now, and I have an integral in my homework that just seems to be impossible to do with the techniques my class has been taught, unless there is a typo somewhere. I have to integrate (2x-3)/((4x-x^2)^1/2). Now, I have successfully integrated this function using integration by parts, as well as trigonometric substitution. However, I am technically not supposed to know how to use those methods. Basically, the only methods I am allowed to use to solve this function are normal substitution and merely knowing a function's antiderivative. So, anyone know how to solve this question with these restrictions?
 October 28th, 2010, 03:43 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Impossible to integrate? Let u = 4x -x^2 then -du = (2x -4)dx Rewrite your numerator as 2x -4+1= (2x-4)dx + 1dx Then you have two integrals... -du/u^(1/2) + 1/something... Not quite sure about the second one. Probably should focus on not getting in a wreck...
 October 28th, 2010, 03:50 PM #3 Newbie   Joined: Oct 2010 Posts: 3 Thanks: 0 Re: Impossible to integrate? That's actually what I did. The first part was easy to solve. However, the second part requires trigonometric substitution. =/
 October 28th, 2010, 03:55 PM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Impossible to integrate? Then our only hope is skipjack, or the ubiquitous counter-example
 October 28th, 2010, 04:50 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 The integral is $\sin^{-1}$$\frac{x-2}{2}$$\,-\,2\sqrt{4x\,-\,x^2}\,+\,\text{C}.$ (Merely knowing an antiderivative is allowed.)
 October 28th, 2010, 05:10 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Impossible to integrate? I would wager a great deal that your knowledge of antiderivatives is a wee bit more extensive than a Calc II student who's never used trigonometric substitutions.
 October 28th, 2010, 08:55 PM #7 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 How so? Most tables of standard integrals include$\int\frac{\,1}{\sqrt{a^2\,-\,x^2}}\,dx\,=\,\sin^{\small{-1}}\frac{x}{a}\,+\,\text{C}$ and it's easy to see that $4x\,-\,x^{\small2}\,=\,2^{\small2}\,-\,(x\,-\,2)^{\small2}.$ It suffices that the original integrand can be rearranged as two terms that are easily recognized as standard forms.
 October 29th, 2010, 07:26 AM #8 Newbie   Joined: Oct 2010 Posts: 3 Thanks: 0 Re: Impossible to integrate? Ah, thanks Slipjack. I actually seem to recall something of that sort from AP Calculus. I am a bit rusty though since I took it two years ago. ;-; I think I remember two others as well, but could you verify them for me? Something about the integral of 1/(((a^2)+(x^2))^(1/2))=arcsec(x/a) and the integral of 1/(((x^2)-(a^2))^(1/2))=arctan(x/a)? I think at least, cannot remember too well. =/
 October 29th, 2010, 04:00 PM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Impossible to integrate? According to the table of integrals in my old calculus textbook: $\int \dfrac{du}{\sqrt{a^2+u^2}}=\ln|u+\sqrt{a^2+u^2}|+C$ $\int \dfrac{du}{\sqrt{u^2-a^2}}=\ln|u+\sqrt{u^2-a^2}|+C$ whereas: $\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\sec^{-1}\left(\dfrac{u}{a}\right)+C$ $\int \dfrac{du}{a^2+u^2}=\dfrac{1}{a}\tan^{-1}\left(\dfrac{u}{a}\right)+C$

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