Hi,

I did this integral this way but the book didn't Convert the left side into arcsin. Am I still right?Attachment 28639Attachment 28640

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- June 17th 2013, 07:12 PMminneola24Two ways to do an integral
Hi,

I did this integral this way but the book didn't Convert the left side into arcsin. Am I still right?Attachment 28639Attachment 28640 - June 17th 2013, 08:07 PMhollywoodRe: Two ways to do an integral
No, I don't think your way works. Before your substitution, you have x in the numerator so it doesn't look like the derivative of an arcsin. After your substitution, you have in the denominator, so again it doesn't look like the derivative of an arcsin.

But the integral is easy enough to do - you just substitute and it becomes a power of w.

- Hollywood - June 17th 2013, 09:06 PMminneola24Re: Two ways to do an integral
Actually on the left side the (x-3) gets crossed out by replacing the dx with du / [2 (x-3) ]

- June 17th 2013, 09:44 PMProve ItRe: Two ways to do an integral
Your substitution looks ok, but you should have ended up with . There is NO point in bringing in extra square roots, as they would require more substitutions to be able to integrate.

What you should do is let and your integral becomes . Surely you can integrate that... - June 17th 2013, 09:55 PMminneola24Re: Two ways to do an integral
Thanks - but why does it require more substituion to integrate that? I obviously see there is an easier way to do the problem - I just want to know if doing 3^2 - (u^.5)^2 is still correct (after putting it into arcsin)

- June 17th 2013, 10:53 PMProve ItRe: Two ways to do an integral
I just told you that it is not, because you will be required to integrate an extra square root, which requires more substitutions!

- June 18th 2013, 06:03 AMminneola24Re: Two ways to do an integral
But there is a square root in arcsin right?

- June 18th 2013, 06:51 AMtopsquarkRe: Two ways to do an integral
What hollywood and Prove It are saying is if it were simply then we can simply integrate. But your form of requires another substitution for the square root over u. It's the inner square root that is the problem here, not the outer.

-Dan - June 18th 2013, 07:34 AMminneola24Re: Two ways to do an integral
I see what you're saying - sorry!

So whatever is in between the parenthesis of (root u)^2 has to be just and only u^2

u cant be anything but u?