Question

Integral of the hour

Answer 1

\[ \int \sqrt{\frac{x}{c-x}} \ dx \]

Answer 2

I already give up LOL

Answer 3

Wolfram gives an unuseful answer. This is another nice triumph for pencil and paper. ;-)

This integral is physically useful. I just used it for this problem:
http://openstudy.com/study#/updates/4f0de661e4b084a815fcff56

Answer 4

Well what is the answer??????????????

Answer 5

Can't keep us in suspense

Answer 6

Oh turing test knows it :D

Answer 7

no, just saying: patience...

Answer 8

Try it.

Answer 9

why do CAS always give weird answers whenever there's a square root involved in the integral?

Answer 10

I am in the midst of it but......... idk

Answer 11

it's easy, just use substitution

Answer 12

which one? I can't find a simple u-sub....

Answer 13

uh oh :O

Answer 14

just substitute c-x by u

Answer 15

I tried that, have you?

Answer 16

nope

Answer 17

me too and i get stuck

Answer 18

If you had you'd know it doesn't work at all

Answer 19

quite right TT

Answer 20

JamesJ is probably gloating at us right now :-D

Answer 21

k just tell us the answer

Answer 22

I tried dividing it out, difference of squares tricks....
trig sub is out of place...

Answer 23

I don't gloat; not my style. I'll tell you the substitution if you like.

Answer 24

shoot :)

Answer 25

All ears

Answer 26

it's probably something really simple

Answer 27

\[ x = c \ \cos^2 t \]

Answer 28

okay... that was nontrivial

Answer 29

how did you come up with that?

Answer 30

a square trig sub?
simple, yet subtle ;-)
very nice.

Answer 31

oh!

Answer 32

No, I just came in. you guys give up too fast -.-

Answer 33

It's like a regular trig sub but sort of making up for the non-squared term under the radical, right James?
Something like that?

Answer 34

well let me try...

Answer 35

yes, exactly

Answer 36

I will try to solve it with another substitution, if possible.

Answer 37

I just came in too :-(

Answer 38

I have got some really nice integrals. I will post them.

Answer 39

I think I have another substitution that would work, a bit lengthy though. Write
\[\sqrt{\frac{x}{c-x}}=\sqrt{-1+\frac{c}{c-x}}\]
then let \(u=\frac{c}{c-x}\). I think this substitution will work, but another substitution (an obvious one) is required.

Answer 40

Yes, you're going to need a trig substitution at some stage.

Answer 41

@ Mr. Math I started with that one
@ James, I have an answer in terms of t, but I don't know if I messed up
Posting in a sec...

Answer 42

Skip writing down that step. Just go to the final answer.

Answer 43

Oh, never mind, messed up already...
trying again...

Answer 44

You're right James, but not necessarily the substitution you suggested.

Answer 45

sure, sin^2 works as well. The only point I'm making is there must be an inverse function in the answer, so there is almost certainly a trig substitution somewhere along the way.

Answer 46

Yeah right. I will need to use a tan substitution;
If I use that substitution, I will get:
\[I=c\int \frac{\sqrt{u-1}}{u^2}du\]
By setting \(t=\sqrt{u-1}\), then
\[I=2c\int \frac{t^2+1-1}{(t^2+1)^2}=2c\int (\frac{1}{t^2+1}-\frac{1}{(t^2+1)^2})dt.\]

Answer 47

The first term is an arctan, substituting \(t=\tan\theta\) would do the other term.

Answer 48

But definitely the substitution \(x=c\cos^2{t}\) is way smarter.