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Famous arcsin formula

  • Thread starter Riazy
  • Start date
  • #1
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$e^x / Sqrt[1-e^[2x]] (An integral with e^x div by algebraic expressions)(Scan paper)

Homework Statement



Ok I will try my best to write this with digital characters: $e^x / Sqrt[1-e^[2x]]
You will see the problem attached on an image below for more details.

e^x is present

Homework Equations



When I do a variable switch, I recalled a famous arcsin formula (also listed below on my scanned paper)


The Attempt at a Solution



See the picture below

[PLAIN]http://img839.imageshack.us/img839/2892/img0003001.jpg [Broken]
 
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Answers and Replies

  • #2
86
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You were right to think that it was an arcsin integral. This is how it is derived:

You have to do substitution it twice and there is a restricting on x.
2eoya8m.jpg
 
  • #3
rock.freak667
Homework Helper
6,230
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If your integral became

[tex]\int \frac{u}{\sqrt{1-u^2}} du [/tex]

then just do another substitution like t=1-u2, which would make your integral much simpler.
 
  • #4
86
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If your integral became

[tex]\int \frac{u}{\sqrt{1-u^2}} du [/tex]

then just do another substitution like t=1-u2, which would make your integral much simpler.
But if you do that, wouldn't there be a radical in the numerator?
If you let t = 1-u2
Then...

[tex]\int \frac{u}{\sqrt{1-u^2}} du [/tex] becomes [tex]\int \frac{\sqrt{t^2-1}}{t} du [/tex]
Or am I missing something?
 
  • #5
rock.freak667
Homework Helper
6,230
31


But if you do that, wouldn't there be a radical in the numerator?
If you let t = 1-u2
Then...

[tex]\int \frac{u}{\sqrt{1-u^2}} du [/tex] becomes [tex]\int \frac{\sqrt{t^2-1}}{t} du [/tex]
Or am I missing something?
You'd need to get dt=2u du, such that u du=dt/2.
 
  • #6
30
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how did you make v = sin x = arcsin (v)

Because as far as I am concerned its not appearing in the integral yet


Wait.. Did you use U as e^x and both sin x at the same time`?, both of them up there are u:s then?
 
Last edited:
  • #7
rock.freak667
Homework Helper
6,230
31


how did you make v = sin x = arcsin (v)

Because as far as I am concerned its not appearing in the integral yet


Wait.. Did you use U as e^x and both sin x at the same time`?, both of them up there are u:s then?
Your integral contains an extra 'u' term in the numerator, so you can't really just use the result to get arcsine.
 
  • #8
108
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Your integral contains an extra 'u' term in the numerator, so you can't really just use the result to get arcsine.
And it should not contain that extra u term.

If you use u=e^x, then du=e^x dx

Your original integrand: e^x dx / (sqrt(1-e^2x))

So you get a new integrand: du / (sqrt (1-u^2)) ... which fits the formula from the back of your head. Planauts basically showed how that arcsin derivative was confirmed.
 
  • #9
rock.freak667
Homework Helper
6,230
31


Next time I should read the entire integral to be done instead of the last one :redface:
 

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