More damn integration >< (1 Viewer)

[Rush152]

int (x^3 (6 + x^2)^3) dx
let u = 6 + x^2

thfor, u' = 2x

simply use substitution eh? the only thing that I don't know how to do is to use substitution when the x at the front has a higher power than u'

i don't think we've ever been taught how to do it =?
and the 3u book (fitzpatrick) that we use only has one example of it in the entire book!

could someone just run thru the process of how to do these type of qs pleaaase?? =)


*edited*

 

[Archman]

expand it.

 

[tempco]

x^2 = u - 6

1/2 int u(u-6).du

 

[smallcattle]

x^2 = u - 6

1/2 int u(u-6).du

its x^3 not x^2

 

[Rorix]

yes but when you use your 2xdx = du it becomes 1/2 x^2(x^2 + 6) du


or you could just like
yknow
expand

 

[Rush152]

ahh yea you could expand it XD
bleh i forgot to add the ^3 at the end there no wounder..

ive fixed the question a bit so you can't expand it and have to use substitution heh, how how how??

 

[tempco]

x^2 = u - 6

1/2 int u^3(u-6).du

1/2 int (u^4 - 6u^3) du

 

[jumb]

Use integration by parts.

Or is that only a 4unit thing?

 

[withoutaface]

Use integration by parts.

Or is that only a 4unit thing?

Int by parts is 4u, yes.

 

[jumb]

you cant sue that in 3unit your stupid fool

Why not? Itll be heaps easier.

 

[jumb]

Actually now that I think about it, It would still be hard. Maybe even harder.

 

[sub]

actually u CAN use 4u techniques in a 3u exam...cos its just another method of doing it...its just a pity that 3u ppl cant benefit from it. good on ya jumb...thats the way i woulda dun this one

 

[withoutaface]

excuse me withoutaface, how come you keep deleting my threads, dont you like me, maybe if you would if you got a chance

Your threads were deleted because they were stupid. Read the forum rules before posting.

kthxbai

 

[Danny11]

actually u CAN use 4u techniques in a 3u exam...cos its just another method of doing it...its just a pity that 3u ppl cant benefit from it. good on ya jumb...thats the way i woulda dun this one

how? so if they tell you to integrate by substition, you use by parts or inspection, you still get the full mark?

 

[jumb]

You SHOULD unless they sepcifically specify otherwise.

But you never know how the marking guidlines work.

 

[Rush152]

mmm... so technically this would be a 4u thing only?

 

[Danny11]

You SHOULD unless they sepcifically specify otherwise.

what? like how they say DON'T USE BY PARTS OR INSPECTION. YOU MUST SUBSTITUTE USING U=XXX.

 

[akira276]

final answer is:
1/2[(u^5)/5 -3/2(u^4)] +c with u=6+x^2

ive checked it so im pretty sure but u can check it agen if u want...btw i did this thru substitution of u=6+x^2 in the original int

 

[Rush152]

nice akira!!

i got the same answer *whew*

thanks all for replyin