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for the life of me I cannot recall how to integrate say
Int(dv/(k-v^2))=dt
Its been 30 years since I took calc, and can't find my Taylor book.
Int(dv/(k-v^2))=dt
Its been 30 years since I took calc, and can't find my Taylor book.
Why is there a dt thingy following the integral expression? So, I assume that you want to integrate:denverdoc said:for the life of me I cannot recall how to integrate say
Int(dv/(k-v^2))=dt
Its been 30 years since I took calc, and can't find my Taylor book.
VietDao29 said:Why is there a dt thingy following the integral expression? So, I assume that you want to integrate:
[tex]\int \frac{dv}{k - v ^ 2}[/tex]
If k > 0, you can do something like this:
[tex]\int \frac{dv}{k - v ^ 2} = \int \frac{dv}{(\sqrt{k} - v) (\sqrt{k} + v)} = \frac{1}{2 \sqrt{k}} \int \frac{(\sqrt{k} - v) + (\sqrt{k} + v)}{(\sqrt{k} - v) (\sqrt{k} + v)} dv[/tex]
[tex]= \frac{1}{2 \sqrt{k}} \int \left( \frac{1}{\sqrt{k} + v} + \frac{1}{\sqrt{k} - v} \right) dv = ...[/tex]
Can you go from here? :)
If k < 0 (or in other words, -k > 0), then you can pull out the minus sign, like this:
[tex]\int \frac{dv}{k - v ^ 2} = - \int \frac{dv}{- k + v ^ 2} = - \int \frac{dv}{(\sqrt{- k}) ^ 2 + v ^ 2} = -\frac{1}{\sqrt{-k}} \arctan \left( \frac{x}{\sqrt{-k}} \right) + C[/tex], can you get this? :)