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

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

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

CallistoBest 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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
looks like i made error
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah i made error ...  ( 1  cos ..) = 
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
we get pretty close to dw:1349076028513:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
this is pretty close from solution of WA
 one year ago

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

CallistoBest ResponseYou've already chosen the best response.1
Why should we express t in terms of arc tan?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
I don't know ... I just want match up our answer with WA
 one year ago

CallistoBest 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*
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Let's graph our solution
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
something is missing
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ \tan t = \sin t / \cos t = \sqrt{\frac{ax}{xb}} \]
 one year ago

CallistoBest 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 :'(
 one year ago

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

experimentXBest 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.
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Oh!!! Let me redo it....
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
also change that value inside arctan it should be \[ \sqrt{xa \over bx}\]
 one year ago

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

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

CallistoBest ResponseYou've already chosen the best response.1
I meant what wolf showed you...
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yep!! http://www.wolframalpha.com/input/?i=Integrate+sqrt%28%28ax%29%28xb%29%29
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
cancel those terms at the front
 one year ago

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

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

experimentXBest ResponseYou've already chosen the best response.1
it seems we are correct!! Wolf is pretty nasty!!
 one year ago

experimentXBest 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!!
 one year ago

CallistoBest 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..
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1349079549133:dw
 one year ago

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

experimentXBest 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.
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
But what about the coefficient??
 one year ago

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

experimentXBest ResponseYou've already chosen the best response.1
Looks like i missed some  somewhere.
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Ehhhhh... I love my solution *v*
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah .. no need to go though rigorous way as Wolf.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
the above is just equally right!!
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
It's always good know to how to get the wolf solution. Thanks !!!
 one year ago

experimentXBest 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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1349081258153:dw
 one year ago

experimentXBest 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)
 one year ago

CallistoBest ResponseYou've already chosen the best response.1
Eh?! three ways to do it?!
 one year ago

experimentXBest 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.
 one year ago

CallistoBest 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 :\
 one year ago
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