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experimentX Group TitleBest ResponseYou've already chosen the best response.1
the answer of the integration is \[ {1 \over 4} \sqrt{b^2 a^2}(ba)\left( t + {\sin (4t)\over 4}\right)\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
On the other hand, Mathematica expresses solution as \[ \frac{\sqrt{ax} \sqrt{b+x} \left(2 \sqrt{ax} \sqrt{b+x} (ab+2 x)(ab)^2 \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right]\right)}{8 \sqrt{(ax) (b+x)}} \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is minus \[ {1 \over 4} \sqrt{b^2 a^2}(ba)\left( t  {\sin (4t)\over 4}\right) \]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Hmm.... \[\int \sqrt{(ax)(xb)}dx\]Let \(x=acos^2t + bsin^2t\) \[dx = (2a\sin t\cos t + 2b \sin t\cos t)dt = (ba)sin2t dt\] Then, the integral becomes (ax) = a  (acos^2 t + bsin^2 t) = a (sin^2 t)  b sin^2 t = (ab) sin^2 t (xb) = (acos^2t + bsin^2t)  b = acos^2 t  b (cos^2 t) = (ab) cos^2t \[\int \sqrt{(ab)sin^2t \times (ab)cos^2t} (ba)sin2tdt\]\[=\int [(ab)sint cost] (ba)sin2tdt\]\[=\frac{1}{2}\int (ba)^2sin^22tdt\] \[=\frac{1}{2}(ba)^2\int sin^2 2tdt\]\[=\frac{1}{2}(ba)^2\int \frac{1cos4t}{2}dt\]\[=\frac{1}{4}(ba)^2 (t\frac{1}{4}sin4t)+C\]Anything wrong here?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
looks like i made error
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah i made error ...  ( 1  cos ..) = 
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
If we put \[ \sin (t) = \sqrt{xa\over ba} \\ \cos (t) = \sqrt{bx\over ba} \\ t = \arctan \sqrt { \left( xa \over bx\right) } \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
we get pretty close to dw:1349076028513:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is pretty close from solution of WA
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Cleaning up the solution i get \[ \left(2 \sqrt{ax} \sqrt{b+x} (ab+2 x)(ab)^2 \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right]\right) \over 8 \]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Why should we express t in terms of arc tan?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
I don't know ... I just want match up our answer with WA
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Perhaps... That help get a prettier t... \[tan t = sint / cost = \sqrt{\frac{xa}{ba}}\] *Still working on that*
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Let's graph our solution
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
something is missing
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ \tan t = \sin t / \cos t = \sqrt{\frac{ax}{xb}} \]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
:'( I got into trouble!!! \[=\frac{1}{4} (ab)^2 [\tan^{1} (\sqrt{\frac{xa}{ba}})\frac{\sqrt{(xa)(bx)}}{ba}(1\frac{xa}{ba})]+C \]\[=\frac{1}{4} (ab)^2 [\tan^{1} (\sqrt{\frac{xa}{ba}})\frac{\sqrt{(xa)(bx)}(bx)}{(ba)^2}]+C \]\[=\frac{1}{4} (ab)^2 [\tan^{1} (\sqrt{\frac{xa}{ba}})]+\frac{1}{4}\sqrt{(xa)(bx)}(bx)+C \]Blah... Doesn't seem right :'(
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
I almost got same result as mathematica except for this one \[ \left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right] \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ =\frac{1}{4} (ab)^2 [\tan^{1} (\sqrt{\frac{xa}{ba}})\frac{\sqrt{(xa)(bx)}}{ba}(12\frac{xa}{ba})]+C \] you missed 2 here.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Oh!!! Let me redo it....
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
also change that value inside arctan it should be \[ \sqrt{xa \over bx}\]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Sorry.... typos!!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Once again, it doesn't look great at all... \[\frac{1}{4}(ab)^2 [ tan^{1} (\sqrt{\frac{xa}{bx}})  \frac{\sqrt{(xa)(bx)}}{ba}[12(\frac{xa}{ba})]] +C\]\[=\frac{1}{4}(ab)^2 [ tan^{1} (\sqrt{\frac{xa}{bx}})  \frac{\sqrt{(xa)(bx)}}{ba}(\frac{a+b2x}{ba})] +C\]\[=\frac{1}{4}(ab)^2 tan^{1} (\sqrt{\frac{xa}{bx}}) + \frac{1}{4}\sqrt{(xa)(bx)}(a+b2x) +C\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
so far so good every part matches solution from WA except \[ \tan^{1} \left(\sqrt{\frac{xa}{bx}}\right ) \]
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
What is that in WA?
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Wolf ...
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
I meant what wolf showed you...
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep!! http://www.wolframalpha.com/input/?i=Integrate+sqrt%28%28ax%29%28xb%29%29
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
cancel those terms at the front
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ \frac{\sqrt{ax} \sqrt{b+x} \left(2 \sqrt{ax} \sqrt{b+x} (ab+2 x)(ab)^2 \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right]\right)}{8 \sqrt{(ax) (b+x)}} \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Cancelling those first terms we get \[ {1 \over 8}\left(2 \sqrt{ax} \sqrt{b+x} (ab+2 x)(ab)^2 \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right]\right)\]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
it seems we are correct!! Wolf is pretty nasty!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
We need to prove that \[ \tan^{1} \left(\sqrt{\frac{xa}{bx}}\right ) = \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right] \] this is not correct ... however I'll show you this is equivalent in integrals!!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
No... not really, the coefficient of arctan (simpler one) is 1/4, while the coefficient of arctan (wolf) is 1/8..
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1349079549133:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
there is a formula that \( \arctan(x) + \arctan(1/x)= {\pi \over 2}\) dw:1349079814250:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
hence we have \[ \tan^{1} \left(\sqrt{\frac{xa}{bx}}\right ) = {\pi \over 2}  \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right]\] But this \( \pi \over 2 \) is constant and you can ignore it. Hence the result of the wolf follows.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
But what about the coefficient??
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
\[ =\frac{1}{4}(ab)^2 \tan^{1} \left(\sqrt{\frac{xa}{bx}}\right ) + \frac{1}{4}\sqrt{(xa)(bx)}(a+b2x) +C \\ \frac{1}{4}(ab)^2 {1\over 2 }\left( {\pi \over 2}  \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right] \right ) + \frac{1}{4}\sqrt{(xa)(bx)}(a+b2x) +C\\ \frac{1}{8}(ab)^2 \left( \text{ArcTan}\left[\frac{a+b2 x}{2 \sqrt{ax} \sqrt{b+x}}\right] \right ) + \frac{1}{4}\sqrt{(xa)(bx)}(a+b2x) \\ +C {\pi \over 16 } (ab)^2\\ \text{ Just let } C {\pi \over 16 } (ab)^2 = C_1 \text{ some constnat} \]
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
Looks like i missed some  somewhere.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Ehhhhh... I love my solution *v*
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah .. no need to go though rigorous way as Wolf.
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the above is just equally right!!
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
It's always good know to how to get the wolf solution. Thanks !!!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
well ... remember this \( \arctan(x) + \arctan(1/x)= {\pi \over 2} \) and after integration you can manipulate your solution by adding or subtracting constants. expanple: \( c + \sin^2 x\) and \( c  \cos ^2 x\) are equivalent solution.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Why.......
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1349081258153:dw
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this is typical when you integrate \[ \int \sin x \cos x \; dx\] one time let sin(x) = u ... another time let cos(x) = u ... another time put sin(x)cos(x) = 1/2 sin(2x)
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Eh?! three ways to do it?!
 2 years ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
if the difference between two solution is constant then they are equivalent. well you could put it up that way. but they were all equivalent.
 2 years ago

Callisto Group TitleBest ResponseYou've already chosen the best response.1
Hmm.. I remember someone told me that when I got a different answer from the textbook, it may not be the case that I did it wrong. sometimes, the answers may be different by some constants.. Perhaps, that's the case :\
 2 years ago
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