Calc II substitution integration

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Homework Help Overview

The discussion revolves around the integration of the function 1/(x^(1/2)(1-x)^(1/2)), with participants exploring substitution methods to simplify the problem. The subject area is calculus, specifically focusing on integration techniques.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various substitution methods, with one suggesting letting u = x^(1/2) and another proposing a different algebraic manipulation. Questions arise about the transformation from the original integral to the form presented in the solution manual.

Discussion Status

Some participants are sharing their thought processes and attempts at the problem, while others are questioning the effectiveness of different approaches. There is an acknowledgment of the complexity involved and a recognition of the need for algebraic proficiency in tackling such problems.

Contextual Notes

Participants express concerns about the foundational algebra skills required for higher-level mathematics, indicating a broader context of educational preparedness in mathematics.

kdinser
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Hi all, I'm totally lost on how to get started on this one.

Integrate 1dx/(x^(1/2)(1-x)^(1/2))

The solution manual jumps right through the substitution and ends up as

Integrate (2u*du)/(u(1-u^2)^(1/2))

I'm just not seeing it, little help please.

Keith
 
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If you want to figure out what the solution manual did, try to identify the similarities between the two formulas, and identify what had to change to turn one into the other.


The solution manual's way isn't the only way to do things anyways... what did you think would be a good thing to try to integrate this?
 
Got it, thanks. My derivative taking was rustier then I thought.

Let u=x^(1/2) which makes x=u^2

here's where I kept screwing up
du = 1/(2(x^(1/2)) dx
so dx = 2 x^(1/2) or, as was declared above 2u. At that point the u's cancel and the 2 moves outside and it's in the form of a standard arcsin.

You mentioned there might be another way to do it, I'm curious as to what that might be.
 
algebra is a B1ach isn't it :-). I wish High school teachers took algebra more seriously because when students hit college, they are woefully unprepared for the amount of algebra they will need when learning higher level mathematics, and I can tell you that when I was in high school, I did not even thing about college level math and how much I would need to know from algebra.
 
Another option is (though somewhat longer) to write the integrandum as 1/(x-x²)^(1/2)

Then write as denominator sqrt(1/4 - (1/2-x)²), this equals sqrt (x - x²). Now just get the 1/4 out of the squareroot in order to come to : 1/2 sqrt(1-2²(1/2 -x)²) and now substitute u = 2(1/2-x) and you get a arcsin...

Ok ? Just for what it is worth

regards
marlon
 
Thanks Marlon, I tried something like that at first, but it got messy so I bailed on it :). It did remind me of a few algebraic tricks I think I used to know about though.
 
But you get what I am trying to say, kdinser?

If so, then i am :biggrin:

regards
marlon
 
modmans2ndcoming said:
algebra is a B1ach isn't it :-). I wish High school teachers took algebra more seriously because when students hit college, they are woefully unprepared for the amount of algebra they will need when learning higher level mathematics, and I can tell you that when I was in high school, I did not even thing about college level math and how much I would need to know from algebra.

word. I hated math in high school. I'm now wishing I could go back and spend less time on S, D, and R&R and more time on understanding complex relationships between numbers.
 

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