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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: trinon on November 06, 2008, 01:44:37 am

Title: HOW TO: Anti derivatives through derivatives.
Post by: trinon on November 06, 2008, 01:44:37 am: Ever have an equation you want to derive, but couldn't because you plain don't know how? Well than this is the guide for you!

As a side note, this is actually covered under the Methods study design and a question like this will most probably be asked on either of the two exams.

So without further Apu (hehe, Simpsons related joke):

TRINON'S GUIDE TO ANTI-DERIVATIVES THROUGH DERIVATIVES

I'm only going to run through the fundamental method, because there isn't much else to it. It only starts getting hard in Specialist Maths when they start throwing things like differentiate $xCos^{-1}(3x)$ and hence anti-differentiate $Cos^{-1}(3x)$ and things like that.

We start off with an equation that we can't anti differentiate with any method that has been covered in the methods study design.

$y = log_e(x)$

We first multiply this equation by $x$ so that we get $xlog_e(x)$ .

Next we find the derivative via the product rule:

$\frac{d}{dx}(xlog_e(x)) = log_e(x) + \frac{x}{x} = log_e(x) + 1$

Next we re-arrange the new equation:

$log_e(x) = \frac{d}{dx}(xlog_e(x)) - 1$

If we now anti-derive both sides we get:

$\int log_e(x) dx = \int\frac{d}{dx}(xlog_e(x)) dx - \int1 dx$

$\implies \int log_e(x) dx = xlog_e(x) - x + C$

Now you can Anti-derive the in-anti-derivable!

Hope this helps guys. If you've got any questions just ask.
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Glockmeister on November 06, 2008, 01:49:12 am: Nice... didn't know about this actually
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Collin Li on November 06, 2008, 01:53:21 am: BTW, you cannot get asked to do something like this unless they provide you with the necessary function to derive first, then hence, using that result, anti-derive it to solve the more difficult integral.
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Mao on November 06, 2008, 02:23:54 am: things like $xe^{x}$ , $x^2e^x$ and other assorted don't work with this. :P
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: trinon on November 06, 2008, 03:03:49 am: Quote from: Mao on November 06, 2008, 02:23:54 am
things like $xe^{x}$ , $x^2e^x$ and other assorted don't work with this. :P

Yeah, we know. This was more for the "Differentiate $x*this$ and hence anti-differentiate $this$ ". It's not a perfect solution..
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: dcc on November 06, 2008, 09:14:15 am: Quote from: Mao on November 06, 2008, 02:23:54 am
things like $xe^{x}$ , $x^2e^x$ and other assorted don't work with this. :P

$\dfrac{d}{dx} \left( xe^x \right) = x e^{x} + e^{x} \implies \int xe^{x} \ dx = xe^{x} - e^x + c_{1}$

$\dfrac{d}{dx} \left(x^2 e^x \right) = 2x e^x + x^2 e^x \implies \int x^2 e^x = x^2e^x - 2 \int x e^x \ dx = e^x \left(x^2 - 2x + 2\right) + c_{2}$

And so on :P
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: onlyfknhuman on November 06, 2008, 10:00:05 am: Holy kcuf genius ! thanks
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: BiG DaN on November 06, 2008, 11:01:22 am: this is just antidifferentiation by recognition yeh?
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: excal on November 06, 2008, 11:14:52 am: Also, you can use \cos{x} to represent $\cos{x}$
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: trinon on November 06, 2008, 11:22:32 am: Quote from: Excalibur on November 06, 2008, 11:14:52 am
Also, you can use \cos{x} to represent $\cos{x}$

I prefer brackets, it gives exactly what you're doing rather than "Hmm, do I include that in the cos or don't I?" dilemma.
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Mao on November 06, 2008, 11:44:08 am: Quote from: trinon on November 06, 2008, 11:22:32 am
Quote from: Excalibur on November 06, 2008, 11:14:52 am
Also, you can use \cos{x} to represent $\cos{x}$

I prefer brackets, it gives exactly what you're doing rather than "Hmm, do I include that in the cos or don't I?" dilemma.

no, as in non-italic $\cos (x)$ and $\log_e (x)$ as opposed to $cos(x),\; log_e(x)$
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: trinon on November 06, 2008, 11:45:50 am: Quote from: Mao on November 06, 2008, 11:44:08 am
Quote from: trinon on November 06, 2008, 11:22:32 am
Quote from: Excalibur on November 06, 2008, 11:14:52 am
Also, you can use \cos{x} to represent $\cos{x}$

I prefer brackets, it gives exactly what you're doing rather than "Hmm, do I include that in the cos or don't I?" dilemma.

no, as in non-italic $\cos (x)$ and $\log_e (x)$ as opposed to $cos(x),\; log_e(x)$

I fail to care either way. Makes no difference to me.
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Cthulhu on November 06, 2008, 11:59:53 am: cos is cos and log is log, who cares? Thanks trinon :D
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: excal on November 06, 2008, 12:03:37 pm: Quote from: trinon on November 06, 2008, 11:22:32 am
Quote from: Excalibur on November 06, 2008, 11:14:52 am
Also, you can use \cos{x} to represent $\cos{x}$

I prefer brackets, it gives exactly what you're doing rather than "Hmm, do I include that in the cos or don't I?" dilemma.

Then \cos{(x)} (as Mao just used).
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Alice on November 06, 2008, 01:36:03 pm: can I ask a question?

find d/dx 2xsin3x and hence find the exact value of anti (x cos3x )dx

upper limit :pai/6 lower limit 0

thank you
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Collin Li on November 06, 2008, 01:46:29 pm: $\frac{d}{dx}\left[2x \sin{(3x)} \right] = 2\sin{(3x)} + 6x\cos{(3x)}$

Integrating both sides with respect to $x$ :

$\implies 2x \sin{(3x)} = \int 2\sin{(3x)}\, dx + \int 6x\cos{(3x)}\, dx$

$\implies \int 6x\cos{(3x)}\, dx = 2x \sin{(3x)} - \int 2\sin{(3x)}\, dx$

$\implies \int 6x\cos{(3x)}\, dx = 2x \sin{(3x)} + \frac{2}{3} \cos{(3x)} + C$

By the fundamental theorem of calculus, if $\int f(x)\, dx = F(x)$ , then $\int_a^b f(x)\, dx = F(b) - F(a)$

Therefore $\int_0^{\frac{\pi}{6}} 6x\cos{(3x)}\, dx = \left[2x \sin{(3x)} + \frac{2}{3} \cos{(3x)}\right]_0^{\frac{\pi}{6}}$

$= \frac{\pi}{3}\sin\left(\frac{\pi}{2}\right) + \frac{2}{3}\cos\left(\frac{\pi}{2}\right) - \left[0 + \frac{2}{3}\cos{(0)}\right]$

$= \frac{\pi}{3} - \frac{2}{3}$

Therefore $\int_0^{\frac{\pi}{6}} x\cos{(3x)}\, dx = \frac{1}{6}\int_0^{\frac{\pi}{6}} 6x\cos{(3x)}\, dx = \frac{\pi}{18} - \frac{1}{9}$
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: lauzy358 on November 08, 2008, 09:12:45 pm: Quote from: BiG DaN on November 06, 2008, 11:01:22 am
this is just antidifferentiation by recognition yeh?

yepp.
& thanks for posting, trinon
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: zzdfa on November 12, 2008, 03:49:03 pm: if you generalize this you get integration by parts =D
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: Glockmeister on November 12, 2008, 04:06:47 pm: yes we do.. i think someone mentioned it to trinon on the night he published this HOW TO guide
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: schnappy on August 04, 2010, 03:56:18 pm: There's a chapter in my text book on this... 'Integration by Recognition'. Is it a common thing in the exams? I don't really 'get it' too well... also seams pointless. Are the VCAA known to ever ask questions like this? Edit: Well I've worked it out after a bit more though... rather than just showing maths it really should be explained better in the OP, imo...
Title: Re: HOW TO: Anti derivatives through derivatives.
Post by: 8039 on October 05, 2010, 03:40:04 am: Can someone explain wth is this? and why multiply by x?