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

let

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

then you have:

1k ∗ ∫dva2 − v2\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:

1k ∗ ∫dva2 − v2\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.
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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.
 
1a2−v2 = 12a ∗ [1a + v + 1a − v]\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]
 
Subhotosh Khan said:
1a2−v2 = 12a ∗ [1a + v + 1a − v]\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]
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Partial fractions. How does this tie into a hyperbolic result?
 
Subhotosh Khan said:
1a2−v2 = 12a ∗ [1a + v + 1a − v]\displaystyle \frac{1}{a^2 - v^2} \ = \ \frac{1}{2a} \ * \ \left [\frac{1}{a\ + \ v} \ + \ \frac{1}{a \ - \ v}\right ]
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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:
1a2−v2 = 12a ∗ [1a + v + 1a − v]\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...

12a ∗ Lna+va−v = t + C\displaystyle \frac{1}{2a} \ * \ Ln \frac{a+v}{a-v} \ = \ t \ + \ C

a+va−v = e2at\displaystyle \frac{a+v}{a-v} \ = \ e^{2at}

av = e2at +1e2at −1\displaystyle \frac{a}{v} \ = \ \frac{e^{2at} \ + 1}{e^{2at} \ - 1}

Now use the fact:

coth(x) = 12(e2x+1e2x−1)\displaystyle coth(x) \ = \ \frac{1}{2}\left(\frac{e^{2x}+1}{e^{2x}-1}\right )
 
Subhotosh Khan said:
warwick said:
[quote="Subhotosh Khan":3txi8wga]1a2−v2 = 12a ∗ [1a + v + 1a − v]\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...

12a ∗ Lna+va−v = t + C\displaystyle \frac{1}{2a} \ * \ Ln \frac{a+v}{a-v} \ = \ t \ + \ C

a+va−v = e2at\displaystyle \frac{a+v}{a-v} \ = \ e^{2at}

av = e2at +1e2at −1\displaystyle \frac{a}{v} \ = \ \frac{e^{2at} \ + 1}{e^{2at} \ - 1}

Now use the fact:

coth(x) = 12(e2x+1e2x−1)\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

ab = xy\displaystyle \frac{a}{b} \ = \ \frac{x}{y}

then

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

If

ab = xy\displaystyle \frac{a}{b} \ = \ \frac{x}{y}

then

a+ba−b = x+yx−y\displaystyle \frac{a+b}{a-b} \ = \ \frac{x+y}{x-y}
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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

ab = xy\displaystyle \frac{a}{b} \ = \ \frac{x}{y}

ab+1 = xy+1\displaystyle \frac{a}{b} + 1\ = \ \frac{x}{y} + 1

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

again

ab−1 = xy−1\displaystyle \frac{a}{b} - 1\ = \ \frac{x}{y} - 1

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

divide (1) by (2) to get

a+ba−b = x+yx−y\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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