# Integral arcsin

• June 5th 2013, 12:33 PM
minneola24
Integral arcsin
They use u=1-x^2

But what is wrong with what I did?
• June 5th 2013, 12:54 PM
Plato
Re: Integral arcsin
Is this the actual problem?

$\int_{\frac{-1}{2}}^0 {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}}$
• June 5th 2013, 12:57 PM
minneola24
Re: Integral arcsin
Quote:

Originally Posted by Plato
Is this the actual problem?

$\int_{\frac{-1}{2}}^0 {\frac{{xdx}}{{\sqrt {1 - {x^2}} }}}$

What is the difference? Aren't they the same.
• June 5th 2013, 01:04 PM
HallsofIvy
Re: Integral arcsin
By taking $v= x^2$, you wind up with $\int \frac{dv}{\sqrt{1- v}}$ NOT $\int\frac{dv}{\sqrt{1- v^2}}dv$

(Plato wasn't saying there was a difference, he was asking you to verify what your problem is.)
• June 5th 2013, 01:35 PM
minneola24
Re: Integral arcsin
Yes he wrote it correctly. The book uses u=1-x^2

I'm just confused because I did it a different way and got a different answer. What did I do wrong?
• June 5th 2013, 01:46 PM
Plato
Re: Integral arcsin
Quote:

Originally Posted by minneola24
Yes he wrote it correctly. The book uses u=1-x^2

I'm just confused because I did it a different way and got a different answer. What did I do wrong?

You can do this without that blasted u-substitution

What is the derivative of $-\sqrt{1-x^2}~?$
• June 5th 2013, 01:49 PM
minneola24
Re: Integral arcsin
I understand you can do it without the u sub I did, but if let's say this is all that came to your mind at the time what did I do wrong in my work?
• June 10th 2013, 02:41 PM
minneola24
Re: Integral arcsin
Anyone have an idea?

Thanks
• June 10th 2013, 06:02 PM
topsquark
Re: Integral arcsin
Quote:

Originally Posted by minneola24
I understand you can do it without the u sub I did, but if let's say this is all that came to your mind at the time what did I do wrong in my work?

You need to put the whole integrand in terms of u. So (avoiding the question of the limits)
$\int \frac{x}{\sqrt{1 - x^2}}~dx = \int \frac{1/2}{\sqrt{1 - u}}~du$

You can't call u $\sqrt{x}$ when you do the u integral.

-Dan