How do I do this integral?

warwick said:
I don't recall doing an integral like this. I see a hyperbolic function in there. I don't recall learning about those either!

http://i111.photobucket.com/albums/n149 ... 1284433180

http://i111.photobucket.com/albums/n149 ... 1284433206
Click to expand...

dv/[g*sin(?)-k*v[sup:3q5vp9rr]2[/sup:3q5vp9rr]] = dt

let

g*sin(?)/k = a[sup:3q5vp9rr]2[/sup:3q5vp9rr]

then you have:

\(\displaystyle \frac{1}{k} \ * \ \int \frac{dv}{a^2 \ - \ v^2}\)

This is a standard integration in "ln" form. That can be converted to hyperbolic form with little algebra.
 
Subhotosh Khan said:
warwick said:
I don't recall doing an integral like this. I see a hyperbolic function in there. I don't recall learning about those either!

http://i111.photobucket.com/albums/n149 ... 1284433180

http://i111.photobucket.com/albums/n149 ... 1284433206
Click to expand...

dv/[g*sin(?)-k*v[sup:3g6u4c57]2[/sup:3g6u4c57]] = dt

let

g*sin(?)/k = a[sup:3g6u4c57]2[/sup:3g6u4c57]

then you have:

\(\displaystyle \frac{1}{k} \ * \ \int \frac{dv}{a^2 \ - \ v^2}\)

This is a standard integration in "ln" form. That can be converted to hyperbolic form with little algebra.
Click to expand...

Crap. I'm not finding how to convert it, and I don't have my calculus textbook with me here at work.
 
Subhotosh Khan said:
What did you get in "ln" form?
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If there were a v in the numerator, I could have an easy u-substitution.

Integral of 1/u = ln u.
 
\(\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]\)
 
Subhotosh Khan said:
\(\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]\)
Click to expand...

Partial fractions. How does this tie into a hyperbolic result?
 
Subhotosh Khan said:
\(\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]\)
Click to expand...

The integral is (1/2a) [ln (a+v) - ln (a-v)]

(1/2a) ln [(a+v) / (a-v)]
 
That's a straightforward integral then. It just involves ln.
 
I may get partial credit. Should I work out this problem using partial fractions or your method? Also, I don't see where you got v^2. That involves a tanh^2 u and is equivalent to 1 - sech^2.
 
warwick said:
Subhotosh Khan said:
\(\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]\)
Click to expand...

The integral is (1/2a) [ln (a+v) - ln (a-v)]

(1/2a) ln [(a+v) / (a-v)]
Click to expand...

\(\displaystyle \frac{1}{2a} \ * \ Ln \frac{a+v}{a-v} \ = \ t \ + \ C\)

\(\displaystyle \frac{a+v}{a-v} \ = \ e^{2at}\)

\(\displaystyle \frac{a}{v} \ = \ \frac{e^{2at} \ + 1}{e^{2at} \ - 1}\)

Now use the fact:

\(\displaystyle coth(x) \ = \ \frac{1}{2}\left(\frac{e^{2x}+1}{e^{2x}-1}\right )\)
 
Subhotosh Khan said:
warwick said:
[quote="Subhotosh Khan":3txi8wga]\(\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]\)
Click to expand...

The integral is (1/2a) [ln (a+v) - ln (a-v)]

(1/2a) ln [(a+v) / (a-v)]
Click to expand...

\(\displaystyle \frac{1}{2a} \ * \ Ln \frac{a+v}{a-v} \ = \ t \ + \ C\)

\(\displaystyle \frac{a+v}{a-v} \ = \ e^{2at}\)

\(\displaystyle \frac{a}{v} \ = \ \frac{e^{2at} \ + 1}{e^{2at} \ - 1}\)

Now use the fact:

\(\displaystyle coth(x) \ = \ \frac{1}{2}\left(\frac{e^{2x}+1}{e^{2x}-1}\right )\)[/quote:3txi8wga]

I'm embarrassed to say this. I'm still stuck on how you go a/v. I played around with the equations a bit but couldn't get it. :oops:
 
A simple corollary of ratios:

If

\(\displaystyle \frac{a}{b} \ = \ \frac{x}{y}\)

then

\(\displaystyle \frac{a+b}{a-b} \ = \ \frac{x+y}{x-y}\)
 
Subhotosh Khan said:
A simple corollary of ratios:

If

\(\displaystyle \frac{a}{b} \ = \ \frac{x}{y}\)

then

\(\displaystyle \frac{a+b}{a-b} \ = \ \frac{x+y}{x-y}\)
Click to expand...

I don't see how you get the simple corollary. I'll have to take your word for it.
 
warwick said:
Subhotosh Khan said:
A simple corollary of ratios:

If

\(\displaystyle \frac{a}{b} \ = \ \frac{x}{y}\)

\(\displaystyle \frac{a}{b} + 1\ = \ \frac{x}{y} + 1\)

\(\displaystyle \frac{a+b}{b} \ = \ \frac{x+y}{y}\)..............................(1)

again

\(\displaystyle \frac{a}{b} - 1\ = \ \frac{x}{y} - 1\)

\(\displaystyle \frac{a-b}{b} \ = \ \frac{x-y}{y}\)..............................(2)

divide (1) by (2) to get

\(\displaystyle \frac{a+b}{a-b} \ = \ \frac{x+y}{x-y}\)

This is taught - generally in Algebra 1.
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I don't see how you get the simple corollary. I'll have to take your word for it.
Click to expand...
 
warwick said:
I took algebra 1 eleven years ago. Cut me some slack. Lol.
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And I took algebra -1 ~51 years ago... no slack here...
 
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