Question

Intergrate sqrt(x/(a-x))

Answer 1

\[\int\limits_{}^{}\sqrt{x/(a-x)}\]

Answer 2

try trigonometric substitution,
let x = a^2 sin^2Q

read theta for Q

Answer 3

and dx = a^2 2 SinQ CosQ

Answer 4

i think it should be
x= a*sin^2Q not a^2 that way you can factor out the a

but yes i believe it will work

Answer 5

yeah..

Answer 6

Set, \( (a-x) = z \implies x=a-z\)

Diff w.r.t x, \( -dx = dz \)


\[ \int \frac x {(a-x)} \; dx = - \int \frac {(a-z)} {z} \; dz \]

Now separate and integrate.

Answer 7

So, with that I render trig substitution tedious and redundant ;)

Answer 8

what about the square root ?

Answer 9

@experimentX Check my work though.

Answer 10

Oh there was a square root? lol I misread non-latex things :P

Answer 11

haha :)

Answer 12

\[\int\limits_{}^{}\sqrt{\frac{x}{a-x}}dx\]

Answer 13

I did experimentX's sub and got a pretty ugly answer which I suspect is wrong. Can't find the paper right now...

Answer 14

looks like i should give a try too ... have been slacking on simple stuff.

Answer 15

I got\[-2a^{5/2}(\cos\theta+\frac13\sin^3\theta)\]and then I stopped working on it
that was with the original sub though with a^2sin..., not asin...

Answer 16

with x=a*sin^2
it should simplify nicely to
\[2a \int\limits_{}^{}\sin^{2} \theta d \theta\]

Answer 17

@henpen Do you know the right answer?

Answer 18

i got something like this
a(arcsin(sqrt(x/a) - 2sqrt(4ax-4x^2)/a)

Answer 19

what the hell ... i was wrong
http://www.wolframalpha.com/input/?i=integrate+sqrt%28x%2F%28x-a%29%29+dx

Answer 20

Hell is right here when we can't solve a problem.

Answer 21

lol..

Answer 22

I'm now so sure I wanna solve it if that's what the answer looks like :P

Answer 23

not so sure*

Answer 24

Turing, wolfram is making it looking ugly.

Answer 25

yeah I know. I've spent many hours comparing my answers to the wolf's for verification, just because it likes to make things look complicated.

Answer 26

@TuringTest what did you get??

Answer 27

I posted what I got with the original sub above, though I didn't undo the trig sub
I gotta leave so I don't think I have time to redo it

Answer 28

without substitution i got

\[a(\theta -\sin(2\theta)/2)\]

Answer 29

now I got dumbcow's answer without the 2 in front...

Answer 30

i got the same with dumb cow i changed sin^2Q = (1-cos1Q)/2 and integrated.

Answer 31

oh wait, I get exactly dumbcow's integral...
and I get your answer...
so... that is probably right, and wolfram has made it hideous somehow...

Answer 32

is there some definition for arctan in terms of ln ?

Answer 33

hm... it would appear not for real numbers, only complex

Answer 34

but there are logs involved.

Answer 35

I have to go :(
I hope to see a good answer when I return

Answer 36

wolfram does some crazy substitutions and then applies partial fractions
i think both answers are equivalent

Answer 37

trig substitution is far more simple in this case

Answer 38

anyway, here is my solution
\[2a \int\limits_{}^{}\sin^{2} \theta d \theta = 2a*\frac{1}{2}(\theta -\sin \theta \cos \theta) = a \theta-a \sin \theta \cos \theta\]
\[\sin \theta = \sqrt{\frac{x}{a}}\]
\[\cos \theta = \sqrt{\frac{a-x}{a}}\]
\[\theta = \sin^{-1} \sqrt{\frac{x}{a}}\]
\[\rightarrow a \sin^{-1} \sqrt{\frac{x}{a}}-\sqrt{ax-x^{2}}+C\]

Answer 39

@dumbcow, sorry, I wasn't following the posts- what did you substitute x as?

Answer 40

x = a (sin x)^2