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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- Jun 17th 2013, 06:12 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 - Jun 17th 2013, 07:07 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 $\displaystyle \sqrt{v}$ 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 $\displaystyle w=9-(x-3)^2$ and it becomes a power of w.

- Hollywood - Jun 17th 2013, 08:06 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) ]

- Jun 17th 2013, 08:44 PMProve ItRe: Two ways to do an integral
Your substitution looks ok, but you should have ended up with $\displaystyle \displaystyle \begin{align*} \frac{1}{2}\int{\frac{dv}{\sqrt{ 9 -v}}} \end{align*}$. 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 $\displaystyle \displaystyle \begin{align*} w = 9 - v \implies dw = -dv \end{align*}$ and your integral becomes $\displaystyle \displaystyle \begin{align*} -\frac{1}{2}\int{\frac{dw}{\sqrt{w}}} = -\frac{1}{2}\int{ w^{ -\frac{1}{2} } } \end{align*}$. Surely you can integrate that... - Jun 17th 2013, 08:55 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)

- Jun 17th 2013, 09:53 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!

- Jun 18th 2013, 05:03 AMminneola24Re: Two ways to do an integral
But there is a square root in arcsin right?

- Jun 18th 2013, 05:51 AMtopsquarkRe: Two ways to do an integral
What hollywood and Prove It are saying is if it were simply $\displaystyle \sqrt{a^2 - u^2}$ then we can simply integrate. But your form of $\displaystyle \sqrt{a^2 - \left ( \sqrt{u} \right )^2$ requires another substitution for the square root over u. It's the inner square root that is the problem here, not the outer.

-Dan - Jun 18th 2013, 06:34 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?