# Famous arcsin formula

$e^x / Sqrt[1-e^[2x]] (An integral with e^x div by algebraic expressions)(Scan paper) ## Homework Statement Ok I will try my best to write this with digital characters:$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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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.

rock.freak667
Homework Helper

$$\int \frac{u}{\sqrt{1-u^2}} du$$

then just do another substitution like t=1-u2, which would make your integral much simpler.

$$\int \frac{u}{\sqrt{1-u^2}} du$$

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...

$$\int \frac{u}{\sqrt{1-u^2}} du$$ becomes $$\int \frac{\sqrt{t^2-1}}{t} du$$
Or am I missing something?

rock.freak667
Homework Helper

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

$$\int \frac{u}{\sqrt{1-u^2}} du$$ becomes $$\int \frac{\sqrt{t^2-1}}{t} du$$
Or am I missing something?

You'd need to get dt=2u du, such that u du=dt/2.

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?

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rock.freak667
Homework Helper

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.

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.

rock.freak667
Homework Helper

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