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#1 2006-01-05 01:42:05

Andorin
Member
Registered: 2005-10-24
Posts: 4

Integration Problem

Howdy folks,

     I'm just running through my maths book and have got stuck on solving this problem. Being... find the value of:

          ∫[1,0] 1/√(9-4t²) dt

      Try as I might I just cannot work out the correct way to integrate this. I've been trying to use this conversion method:

          f'(x) = 1/√(a²-t²)   ≈   f(x) = sin-¹(t/a)+c

       But I don't get the right answer according to the book no matter what I do. The book says that the value should be 0.365, but I just don't seem to hit the jackpot no matter how I arrange it. I'm assuming it has something to do with the co-efficient of 4 but hey, who knows. Hopefully one of you do and can explain where I'm going wrong..

Cheers in advance

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#2 2006-01-05 01:54:39

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integration Problem

ArcSin[2/3]/2


IPBLE:  Increasing Performance By Lowering Expectations.

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#3 2006-01-05 01:55:43

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integration Problem

≈0.364863828113483181727398329907...


IPBLE:  Increasing Performance By Lowering Expectations.

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#4 2006-01-05 02:13:39

Andorin
Member
Registered: 2005-10-24
Posts: 4

Re: Integration Problem

Cheers the quick response.

Where did the numbers come from though?

Is the two there because (2t)² = 4t², and thus 2t was actually the correct "t" to go there... and is it all divided by two because it was actually two variables being used instead of the intended one?

Last edited by Andorin (2006-01-05 02:22:06)

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#5 2006-01-05 02:47:10

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Integration Problem

" f'(x) = 1/√(a²-t²)   ≈   f(x) = sin-¹(t/a)+c "

This is not right.
1. not f(x), f(t).
2. f'(x)=1/sq(a^2-t^2) => f(t) = ArcSin[t/a]/2+c


IPBLE:  Increasing Performance By Lowering Expectations.

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