# Thread: Two ways to do an integral

1. ## Two 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?

2. ## Re: 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

3. ## Re: 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) ]

4. ## Re: Two ways to do an integral

Originally Posted by minneola24
Actually on the left side the (x-3) gets crossed out by replacing the dx with du / [2 (x-3) ]
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...

5. ## Re: 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)

6. ## Re: 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!

7. ## Re: Two ways to do an integral

But there is a square root in arcsin right?

8. ## Re: 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

9. ## Re: 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?