1. ## Integral arcsin

They use u=1-x^2

But what is wrong with what I did?

2. ## Re: Integral arcsin

Is this the actual problem?

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

3. ## Re: Integral arcsin

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.

4. ## 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.)

5. ## 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?

6. ## Re: Integral arcsin

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 $-\sqrt{1-x^2}~?$

7. ## 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?

8. ## Re: Integral arcsin

Anyone have an idea?

Thanks

9. ## Re: Integral arcsin

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