- #1

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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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- Thread starter kdinser
- Start date

- #1

- 337

- 0

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

- #2

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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The solution manual's way isn't the only way to do things anyways... what did

- #3

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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.

- #4

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- #5

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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

- #6

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- #7

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But you get what I am trying to say, kdinser???

If so, then i am

regards

marlon

If so, then i am

regards

marlon

- #8

Pythagorean

Gold Member

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modmans2ndcoming said:

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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