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experimentX
 3 years ago
Integration:
experimentX
 3 years ago
Integration:

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

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

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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm.... \[\int \sqrt{(ax)(xb)}dx\]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?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1looks like i made error

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

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

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

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

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

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Let's graph our solution

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1something is missing

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

Callisto
 3 years ago
Best 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 :'(

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

experimentX
 3 years ago
Best 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Oh!!! Let me redo it....

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

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

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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1I meant what wolf showed you...

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1cancel those terms at the front

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1We 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!!

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349079549133:dw

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1hence 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.

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1But what about the coefficient??

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Looks like i missed some  somewhere.

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1the above is just equally right!!

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1well ... 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1349081258153:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this 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)

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

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

Callisto
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm.. 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 :\
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