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Different Method of Integration (2 Viewers)

Mill

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I recall someone posting a link to a document about a different method of integration but I have forgotten the 'name' of this method. I was hoping someone would be able to provide a link or some information.

An example of how it works:

Integral of tan x . sec^2 x dx

= Integral of tan x . d(tanx)

= 1/2 . tan^2x + c

ie. you change the question so you are integrating with respect to tan x, as opposed to x, in this example.


Thanks in advance if anyone can help.
 

Riviet

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If I remember right, it's called "direct integration". I had a look around the forums but couldn't find the thread with the link that the guy posted.
 
Last edited:

Templar

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It could also be a substitution, with the choice so obvious you don't bother with the full working out.
 

Mountain.Dew

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its merely integration by observation. in the example provided:

1) Integral of tan x . sec^2 x dx

2) = Integral of tan x . d(tanx)

3) = 1/2 . tan^2x + c


usually, most people would simply go from line 1 to 3 straight away. line 2 is the 'justification' or the reasoning behind going from 1 to 3. its a neat and elegant way to show to you, and the marker, that ur not pulling random figures and algebra out of ur head when integrating.

what i mean by 'integration by observation' - in effect, your brain is thinking: what function, when i integrate it, would give me the function im integration now? --> this method is pretty much how you would integrate logs and exponentials. the 'reasoning' behind em goes like this:

e.g.
Integral of (2x + 1) / (x^2 + x + 6)
= integral of d(x^2 + x + 6) * 1 / (
x^2 + x + 6)
= ln(x^2 + x + 6) + c


note: compare the 2nd line with:

integral of d(@) * 1/(@) -- simply substitute @ with (x^2 + x + 6), and it becomes clearer as to how we got a log answer.

hope it helps, M.D.
 

Mill

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You're all pretty much correct.

But I'm still fairly sure there is a "name" for this method (I tried researching under the name Riviet suggested) and I know there was a document here at some stage about it.

If you ask me, the method is uncommon and should be avoided. Most teachers would probably not understand what you have done.

But I still must continue my search to find that document!

Thanks to all who replied.
 

Rax

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My Friend has tried to create a different method of Integration, rather than Integration by Parts

Supposedly It works

I will post if up if It gets interesting
 

A l

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Mill said:
Integral of tan x . sec^2 x dx

= Integral of tan x . d(tanx)

= 1/2 . tan^2x + c
The way I would do that question is similar (but apparently not the same) to yours. I think "Mountain.Dew" might have mentioned it above. I personally call it a "reverse chain rule" method. If a function of x (raised to any power) is multiplied by its derivative, then the integral can be found quite quickly without the need for substitution:

Notice how sec²x is the derivative of tan x, and tan x is to the power of 1, therefore the integral must be tan x to the power of 2. Also, note that:
d/dx (tan²x) = 2(tan x)(sec²x)
Now that result is double of what we want to inegrate so we halve it, hence:
∫(tan x. sec²x)dx = (1/2)∫2(tan x)(sec²x)dx
[i.e. (1/2)∫2(tan x)(sec²x)dx = (1/2)∫2(tan x)(d/dx [tan x]) dx]
.: ∫(tan x. sec²x)dx = (tan²x)/2 + c

Similarly in the case of ∫(tan²x. sec²x)dx
Note that sec²x is the derivative of tan x, and tan x is to the power of 2, therefore the integral must be tan x to the power of 2. Also note that:
d/dx (tan³x) = 3(tan x)²(sec²x)
Now that result is triple of what we want to inegrate so we take a third of it, hence:
∫(tan²x. sec²x)dx = (1/3) ∫3(tan²x)(sec²x)dx
[i.e. (1/3)∫3(tan²x)(sec²x)dx = (1/3)∫3(tan x)²(d/dx [tan x]) dx]
∫(tan²x. sec²x)dx = (tan³x)/3 + c

This is simply observing the the chain rule in reverse and is a mostly a mental method with little working involved. It only works if a function of x raised to any power is multiplied by its derivative.
I'm pretty sure lots of people here use this method, but I think most resort to substitution for this kind of question.
 
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pLuvia

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My teacher also called it the reverse chain rule, but this question you can do it by observation
 

boon

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pLuvia said:
My teacher also called it the reverse chain rule, but this question you can do it by observation

yeah that what they call
where i come from
far far away so it must be right
serious
reverse chain intergation
 

STx

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Yeh, its sort of a reverse chain rule:

∫[f'(x) . f(x)^n] dx = f(x)^n+1/n+1 + C
 

acullen

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I always called that Integration by Substitution.

∫tan(x)·sec2(x)·dx

Let u=tan(x)
then du=sec2(x)·dx

i.e. the integrand is of the form:
∫u·du
=½u2 + c

As u=tan(x)

∫tan(x)·sec2(x)·dx
=½tan2(x) + c
Where c is an arbitrary constant
 

Riviet

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That's how I would set it out in an exam [just like acullen has shown]. Of course, after you do alot of problems, you could starting figuring it out in your head before putting pen to paper.
 

A l

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acullen said:
I always called that Integration by Substitution.

∫tan(x)·sec2(x)·dx

Let u=tan(x)
then du=sec2(x)·dx

i.e. the integrand is of the form:
∫u·du
=½u2 + c

As u=tan(x)

∫tan(x)·sec2(x)·dx
=½tan2(x) + c
Where c is an arbitrary constant
The reverse chain rule and integration by substitution are different methods. Substitution involves letting a certain function of x be a function of u, which in effect simplifies the integration. The reverse chain rule method involves immediate recognition of a derivative and immediately knowing the primitive function of that derivative.
There are of course limitations to each method. The reverse chain rule method has more limitations than the substitution method because it only works if a function raised to any power is multiplied by its derivative, whereas the substitution method can be used much more widely especially in t-formulae and trigonometric substitutions.
 

acullen

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A l said:
The reverse chain rule and integration by substitution are different methods. Substitution involves letting a certain function of x be a function of u, which in effect simplifies the integration. The reverse chain rule method involves immediate recognition of a derivative and immediately knowing the primitive function of that derivative.
There are of course limitations to each method. The reverse chain rule method has more limitations than the substitution method because it only works if a function raised to any power is multiplied by its derivative, whereas the substitution method can be used much more widely especially in t-formulae and trigonometric substitutions.
This "reverse chain rule" is just the most simplistic case of integration by substitution, there is really no difference in the mathematics, just in the analytical approach.
 

Mill

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I didn't realise I would spark such discussion!


I was only looking for reference!


Anyone who would actually consider using this method is out of their mind. :p It's cumbersome and unknown.


The giant, very loud klaxons are already sounding in my head at the thought of anyone using this method.


Besides, I finished my HSC in 2002. :p


But carry on your discussion if you must!
 

acullen

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Mill said:
Besides, I finished my HSC in 2002. :p


But carry on your discussion if you must!
Just wondering, what are you doing now? You would have just have finished or be finishing a degree at the moment?

I did my HSC back in '04. Those were the days...
 

Mill

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In the fourth year of a Commerce/Science degree at UNSW which should be the last year but due to work commitments (a tutoring college of all places) I'm doing part-time this year and will be completing my degree in 2007 instead.
 

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