Question

What's the integral of: sqrt(8x-x^2)?

Answer 1

i think this is a set up for completing the square

Answer 2

I've already did that, but I cannot come to the final answer at all.

Answer 3

done*

Answer 4

you got to
\[\int \sqrt{16-(x-4)^2}dx\]right?

Answer 5

Yep! But I can't get to the final part. Used trig substitution and everything.

Answer 6

one thing you could do is cheat and look in the back of your book, where no doubt it has a formula for
\[\int \sqrt{a^2-u^2}du\]

Answer 7

the other way i guess it to make the substitution \(u=4\sin(\theta)\)

Answer 8

that is after you already make the mental substitution \(u=x-1\) with \(du=dx\) so there is nothing really to change

Answer 9

This is what I got for the answer. \[8\arcsin((x-4)/4) + (x-4)\sqrt{8x-x^2}\]
Also, this is online, so no book.

Answer 10

ok lets see what we get

Answer 11

\[u=4\sin(\theta), du = 4\cos(\theta)d\theta\] and so after the magic disapperance of the radical you get
\[\int 16\cos^2(\theta)d\theta\]

Answer 12

Yep.

Answer 13

out pops the 16 and you rewrite this mess at
\[8\int (\cos(2\theta)+1)d\theta\]

Answer 14

Hold up. Why not just do integration by parts on the cos^2?

Answer 15

how is that going to work?
well, it might work, but you are going to get a mess i think
standard way to integrate cosine squared is like that

Answer 16

Alright. We weren't taught it like that, so please, continue.

Answer 17

we can try that if you like
you are going to do one of those integrals around in a circle i think
way too much work

Answer 18

in any case if we do this the answer becomes
\[4\sin(2\theta)+8\theta\]

Answer 19

then put \(\theta=\sin^{-1}(\frac{x-4}{4})\) to get the answer i think

Answer 20

Ok. I'm gonna try that real quick.

Answer 21

actually your answer looks good to me in any case

Answer 22

same answer i think that i get , matter of fact i am almost sure of it

Answer 23

Odd. It's saying that it's not the correct answer, but that there's always more than one possibility.

Answer 24

yeah i hate that nonsense
you can probably write this in a raft of different ways, all of which are the same, or that vary by a constant

Answer 25

Alright. Well, honestly, I like your way of doing it much, much better. Far less clutter and BS to deal with.

Answer 26

i am going to make a bet
my best is that if you put \(\int \sqrt{8x-x^2}dx\) in to wolfram, and then put in \(\int \sqrt{16-(x-4)^2}dx\) you get two different answers, even though they are identical

Answer 27

I'll try that now.

Answer 28

Online math is the worst.

Answer 29

http://www.wolframalpha.com/input/?i=sqrt%288x-x^2%29

Answer 30

Alright. Just got it. I rewrote a previous way I did it, just to see, and it accepted it. \[8\arcsin((x-4)/4) + ((2x - 8) / 4)\sqrt(8x-x^2) + c\]

Answer 31

don't you love on line courses?

Answer 32

Oh, they truly are a blessing to college students everywhere.

Answer 33

mylabsplus
webassign
thinkwell?

Answer 34

WebWork, which I think is the same as WebAssign.

Answer 35

have fun

Answer 36

Thanks again!

Answer 37

yw