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integration (2 Viewers)

hatty

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hey


im doing integration by substitution and parts

my problem is, how do u no when to subistitute, or when to do it by parts?
is there a clue u can see or is it just practice and u'll identify shit easier.?

cheers.
 

Affinity

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Do about 300 integration pratice problems.. you will eventually get the feel. :D

involving squares a square roots -> some inverse trig sub.

fractions -> probably some log substitution
 
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Basically anything with (a^2 - x^2)^n, (a^2 + x^2)^n or (x^2 - a^2)^n is trig substition, t=tan(x/2) for anything where the trig can't be easily eliminated - other substitions are pretty obvious where you have some varient of f'(x)f(x) or f'(x)/f(x).

Parts for products that can't be eliminated or for reduction, of course. You should get a feel of it pretty quickly.
 

CM_Tutor

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Would any of the current students like to suggest a suitable substitution to use for integrating x<sup>6</sup> / (1 + x)<sup>8</sup>?

More advanced people, have a go by all means, but please don't spoil the question if you happen to know the answer. Thanks.
 
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CM_Tutor

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Hatty, u = x + 1 will leave you with a form requiring you to expand a degree 6 polynomial, integrate termwise, and then attempt to factorise. It's possible - so kudos for suggesting a way - but there are much better approaches.

nike33 - what does "nvm" mean?
 

+Po1ntDeXt3r+

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DAMMIT I WANT TO ANSWER IT and find out if i was rite.. :*(
oh well .. Cmon 04's..

hmm mabbe i could change 03 to 04 in my sig and FOOL U ALL *evil laugh*
 

CM_Tutor

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Originally posted by +Po1ntDeXt3r+
DAMMIT I WANT TO ANSWER IT and find out if i was rite.. :*(
oh well .. Cmon 04's..

hmm mabbe i could change 03 to 04 in my sig and FOOL U ALL *evil laugh*
You could always PM me, if you really want to know... :)
 

Faera

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uhmm...

would this work?:

subst x = [tanu]^2

and then after simplifying, subs v = sinu

?
 

CM_Tutor

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Originally posted by Faera
uhmm...

would this work?:

subst x = [tanu]^2

and then after simplifying, subs v = sinu
Yes, it will, and it's a much better way to go. Can you figure out the answer that results, in terms of x?

PS: There is still a better substiution - much more elegant, but much harder to see ...
 

nike33

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umm by any chance is the answer [(x-1)<sup>5</sup>(-x+7)] / x<sup>7</sup>

iif this isnt the right answer can you tell us? or pm me? :) thx
 

CM_Tutor

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Originally posted by nike33
umm by any chance is the answer [(x-1)<sup>5</sup>(-x+7)] / x<sup>7</sup>
No, it isn't. The answer is x<sup>7</sup> / 7(1 + x)<sup>7</sup> + C, for some constant C
 

CM_Tutor

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In case anyone was wondering, the best method is to note first that x<sup>6</sup> / (1 + x)<sup>8</sup> = 1 / x<sup>2</sup>(1 + 1 / x)<sup>8</sup>, and then use the substitution u = 1 + 1 / x :)

PS: I know this is hard to see, but my 4u teacher threw it at my class at school, so I figure it's OK for anyone to have a go.
 

Grey Council

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Originally posted by CM_Tutor
In case anyone was wondering, the best method is to note first that x<sup>6</sup> / (1 + x)<sup>8</sup> = 1 / x<sup>2</sup>(1 + 1 / x)<sup>8</sup>, and then use the substitution u = 1 + 1 / x :)

PS: I know this is hard to see, but my 4u teacher threw it at my class at school, so I figure it's OK for anyone to have a go.
uh, thats like impossible to see. Literally.
all right, Keypad, answer honestly, would you have seen it if someone had given it to you last year? Honestly, now... Cause if you wouldn't have seen it, then I can forget about it. mmm
:)
 

CM_Tutor

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Originally posted by Grey Council
uh, thats like impossible to see. Literally.
No, it isn't. As I said above, my class solved it - we spent some while discussing it, but both the u = 1 + 1 / x and
x = tan<sup>2</sup>@ substitutions were suggested after we had discussed the more obvious alternatives, like u = 1 + x, etc. I'm not saying it's easy - it isn't - but it is possible.
 

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