# Find my error :/

• February 7th 2008, 11:08 AM
angel.white
Find my error :/
Can anyone help me figure out what I'm doing wrong:

$\int x^3\sqrt{9-x^2} dx$

---
> Substitute:
> $x=3sin\theta$
> $dx=3cos\theta d\theta$
> $\theta = arcsin(x/3)$
> $cos(\theta)=cos(arcsin(x/3))= \frac{\sqrt{9-x^2}}3$
---

= $\int (3sin\theta)^3\sqrt{9-(3sin\theta)^2}(3cos\theta) d\theta$

= $3^4\int sin^3\theta\cos\theta\sqrt{3^2-3^2sin^2\theta} d\theta$

= $3^4\int sin^3\theta\cos\theta\sqrt{3^2(1-sin^2\theta)} d\theta$

= $3^4\int sin^3\theta\cos\theta\sqrt{3^2cos^2\theta} d\theta$

= $3^5\int sin^3\theta\cos^2\theta d\theta$

= $3^5\int sin\theta (1-cos^2\theta)\cos^2\theta d\theta$

---
> Substitute:
> $u = cos\theta$
> $-du = sin\theta d\theta$
---

= $-3^5\int (1-u^2)u^2 du$

= $-3^5\int (u^2-u^4) du$

= $-3^5(\frac13 u^3 - \frac15 u^5)+C$

= $-3^5(\frac13 cos^3\theta - \frac15 cos^5\theta)+C$

= $-3^5\left(\frac13 \left(\frac{\sqrt{9-x^2}}{3}\right)^3 - \frac15 \left(\frac{\sqrt{9-x^2}}{3}\right)^5\right)+C$

= $-3^5\left(\frac1{3^4} (9-x^2)^{3/2} - \frac1{5*3^5} (9-x^2)^{5/2}\right)+C$

= $-3 (9-x^2)^{3/2} + \frac15 (9-x^2)^{5/2}+C$

---------------------
I've done it several times and gotten the same answer, unfortunately it's an even problem so I can't check my book, but Function calculator says the answer should be:
$-\frac15 x^2(9-x^2)^{3/2}-\frac65(9-x^2)^{3/2}+C$
• February 7th 2008, 11:11 AM
Krizalid
I'm too lazy to check your solution, but you can get another faster one by setting $u^2=9-x^2.$
• February 7th 2008, 11:58 AM
angel.white
Quote:

Originally Posted by Krizalid
I'm too lazy to check your solution, but you can get another faster one by setting $u^2=9-x^2.$

Wow, that was so elegant! (I was so focused on using the method just taught to us that I didn't even consider alternatives)

I used your method, and it came out the same as mine, I'll assume the function calculator's answer was the same as mine, and when I checked it that I checked it incorrectly. (if not that, then I did your way wrong in the same manner I did my way wrong, and if not that, then the function calculator is wrong, and if not that, then math is broken)
• February 7th 2008, 12:06 PM
Krizalid
Quote:

Originally Posted by angel.white
Wow, that was so elegant!

It's not elegant, it's just a faster way to avoid boring trig. calculations and the square root.

Quote:

Originally Posted by angel.white
I used your method, and it came out the same as mine, I'll assume the function calculator's answer was the same as mine, and when I checked it that I checked it incorrectly.

I suggest when you want to verify if your answer is correct or not, see solving derivatives step-by-step. If your result does not match with the software, then you're messing with some calculation. (Of course, software may fail.)

Quote:

Originally Posted by angel.white
(if not that, then I did your way wrong in the same manner I did my way wrong, and if not that, then the function calculator is wrong, and if not that, then math is broken)

Math can't be broken :D
• February 7th 2008, 12:17 PM
colby2152
Quote:

Originally Posted by Krizalid
I'm too lazy to check your solution, but you can get another faster one by setting $u^2=9-x^2.$

That is what I did, and I got an answer similar to Angel.White's.

$u=9-x^2$

$du-2xdx$

$\int x^3 \sqrt{9-x^2}dx = \int \frac{u-9}{2}\sqrt{u}du$

$=\int -4.5u^{0.5}+0.5u^{1.5}du$

$= -3u^{1.5}+0.2u^{2.5}+C$

$= -3(9-x^2)^{1.5}+0.2(9-x^2)^{2.5}+C$
• February 7th 2008, 12:52 PM
angel.white
Quote:

Originally Posted by Krizalid
I suggest when you want to verify if your answer is correct or not, see solving derivatives step-by-step. If your result does not match with the software, then you're messing with some calculation. (Of course, software may fail.)

Oh wow! That would have saved me hours and hours last semester. I remember when I was first learning derivatives, and I got confused about the power rule, and hadn't found all the resources that I have now. And I spent this weekend doing the same problems over and over and over, like each problem 40 times, and comparing my answers from each time to see which ones I came up with the most to decide which ones I thought were most likely to be correct.
Quote:

Originally Posted by Krizalid
Math can't be broken :D

Maybe "broken" isn't an appropriate term, just expressing the idea of an unknown variable which causes math to be different than we understand it to be. (Note that I understand the unlikelihood of this possibility, which is why I placed it at the end of the list of possibilities.)
Quote:

Originally Posted by colby2152
That is what I did, and I got an answer similar to Angel.White's.

Thank you, I've been doing these problems all day each day all week, my head is swimming, so nice to have objective verification.
• February 7th 2008, 01:49 PM
mr fantastic
Quote:

Originally Posted by colby2152
That is what I did, and I got an answer similar to Angel.White's.

$u=9-x^2$

$du$= $-2xdx$

$\int x^3 \sqrt{9-x^2}dx =$ - $\int \frac{9-u}{2}\sqrt{u}du$

$=\int 4.5u^{0.5}-0.5u^{1.5}du$

$= 3u^{1.5}-0.2u^{2.5}+C$

$= 3(9-x^2)^{1.5}-0.2(9-x^2)^{2.5}+C$

I'll just put my 2 cents worth in here (1 cent for each point):

1. Colby, you dropped a minus - it's in the edit and needs to be carried through to the end.

2. The answer of $-3(9-x^2)^{1.5}+0.2(9-x^2)^{2.5}+C$ can be simplified by taking out a common factor of $(9 - x^2)^{1.5}$:

$(9-x^2)^{1.5} \left( -3 + 0.2 [9-x^2] \right) + C$

$= (9-x^2)^{1.5} (-3 + 1.8 - 0.2 x^2) + C$

$= (9-x^2)^{1.5} (-1.2 - 0.2 x^2) + C$

$= -(9-x^2)^{1.5} \left (\frac{6}{5} + \frac{1}{5} x^2 \right) + C$

$= -\frac{1}{5} (9-x^2)^{1.5} (6 + x^2) + C$.

Some obvious variant of this final answer is what a CAS is likely to spit out. And indeed what I'd expect to see as the correct option if it was a multiple choice question ..... And indeed, the answer in the back of the book (if the book gave answers to even numbered questions as well)
• February 8th 2008, 04:44 AM
colby2152
Thanks Mr Fantastic... so it looks like the Function Calculator that Angel was using is wrong. All of us agree on the answer, but disagree with the calculator he used.
• February 8th 2008, 12:54 PM
mr fantastic
Quote:

Originally Posted by colby2152
Thanks Mr Fantastic... so it looks like the Function Calculator that Angel was using is wrong. All of us agree on the answer, but disagree with the calculator he used.

To seal the deal: