Int. Help (1 Viewer)

[sasquatch]

Integrate (16 - x²)/x²

Thanks in advance!

 

[Riviet]

Very easy mate! Just split up the fraction and voila.

P.S in future, look for any obvious/simple methods before considering more advanced/harder methods.

 

[who_loves_maths]

Hi sasquatch,

(16 - x²)/x² = 16/x² - 1

Therefore:
∫(16 - x²)/x² dx = ∫16/x² dx - ∫1 dx

= -16/x - x + C

Hope that helps.


P.S. Riviet how can you just jump to the conclusion that he's looking for a more advanced technique? Maybe there are people out there who are genuinely looking help on the " very easy" (to you) questions?

 

[sasquatch]

hahahah woops i missed something out..

:| i read your post and i thought "what the hell is he talking about"

and yeah

i meant to write Integrate sqrt(16 - x²)/x²

Sorry for wasting both your time rivet and wholovesmaths

 

[Riviet]

P.S. Riviet how can you just jump to the conclusion that he's looking for a more advanced technique? Maybe there are people out there who are genuinely looking help on the " very easy" (to you) questions?

Well I must say, the original integral asked was fairly straight forward, even for a 2 unit student, of course it turned out to be a typo, but I was just reminding sasquatch about checking the easier methods first.
Anyway, now it's not so simple.

 

[sasquatch]

Well I must say, the original integral asked was fairly straight forward, even for a 2 unit student, of course it turned out to be a typo, but I was just reminding sasquatch about checking the easier methods first.
Anyway, now it's not so simple.

Yeah hehe i agree with Rivet. If i couldnt do the question i originally posted..id be stupid to think that i could do 4 unit maths. But yeah..

 

[who_loves_maths]

i meant to write Integrate sqrt(16 - x²)/x²

∫Sqrt(16 - x²)/x² dx

Let x = 4Sin(u) , dx = 4Cos(u) du -----> u = ArcSin(x/4)

I = (1/4)∫Cos²u/Sin²u du

= (1/4)∫Cot²u du

= (1/4)∫(Csc²u - 1) du

Note that d(Cot(u))/du = -Csc²u in much the similar way that d(Tan(u))/du = Sec²u
[It's a good idea to learn these differentials off by heart btw.]

---> I = (1/4)(-Cot(u) - u) + C

= -(1/4)Cot(ArcSin(x/4)) - (1/4)ArcSin(x/4) + C


Hope that helps.


EDIT: Note that Cot(ArcSin(x/4)) = Sqrt(16 - x²)/x

Thus, I = -(1/4)(Sqrt(16 - x²)/x) - (1/4)ArcSin(x/4) + C


EDIT 2: as per Riviet's correction below.

 

[sasquatch]

damn theres too much too remember

Thanks though.

 

[who_loves_maths]

Originally Posted by sasquatch

Originally Posted by Riviet
Well I must say, the original integral asked was fairly straight forward, even for a 2 unit student, of course it turned out to be a typo, but I was just reminding sasquatch about checking the easier methods first.
Anyway, now it's not so simple.

Yeah hehe i agree with Rivet. If i couldnt do the question i originally posted... id be stupid to think that i could do 4 unit maths. But yeah...

People sometimes have mental-blocks which may be attributed to many factors such as deprivation of sleep, general fatigue, or bouts of depression, etc...

Ergo, I don't like to jump to the conclusion that a 4unit student temporarily unable to complete a 2unit mathematics problem must imply an implicit stupidity of that particular individual.

 

[sasquatch]

yeah i guess so..

 

[Riviet]

Just a correction with the substitution step:
∫sqrt(16 - x²)/x² dx = ∫sqrt(16-16sin²u).4dosu du/(16sin²u)
= ∫4cosu.4cosu du/16sin²u
= ∫cot²u du

 

[who_loves_maths]

^ Thanks Riviet.
Must be that mental-block I was talking about huh ...

 

[Pianpupodoel]

Cant be bothered opening a new topic...

but can anyone integrate:

4/[x{(16-x)^1/2}]

 

[onebytwo]

try using a substitution of u=16-x

 

[Riviet]

I think a better substitution would be u=(16-x)^1/2
Then du=-1/[2(16-x)^1/2] and x=16-u²
After making the substitution, you'll probably need to use partial fractions.