Here's the question you clicked on:
experimentX
Integration:
the answer of the integration is \[ {1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t + {\sin (4t)\over 4}\right)\]
On the other hand, Mathematica expresses solution as \[ \frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}} \]
this is minus \[ {1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t - {\sin (4t)\over 4}\right) \]
Hmm.... \[\int \sqrt{(a-x)(x-b)}dx\]Let \(x=acos^2t + bsin^2t\) \[dx = (-2a\sin t\cos t + 2b \sin t\cos t)dt = (b-a)sin2t dt\] Then, the integral becomes (a-x) = a - (acos^2 t + bsin^2 t) = a (sin^2 t) - b sin^2 t = (a-b) sin^2 t (x-b) = (acos^2t + bsin^2t) - b = acos^2 t - b (cos^2 t) = (a-b) cos^2t \[\int \sqrt{(a-b)sin^2t \times (a-b)cos^2t} (b-a)sin2tdt\]\[=\int [(a-b)sint cost] (b-a)sin2tdt\]\[=-\frac{1}{2}\int (b-a)^2sin^22tdt\] \[=-\frac{1}{2}(b-a)^2\int sin^2 2tdt\]\[=-\frac{1}{2}(b-a)^2\int \frac{1-cos4t}{2}dt\]\[=-\frac{1}{4}(b-a)^2 (t-\frac{1}{4}sin4t)+C\]Anything wrong here?
looks like i made error
yeah i made error ... - ( 1 - cos ..) = -
If we put \[ \sin (t) = \sqrt{x-a\over b-a} \\ \cos (t) = \sqrt{b-x\over b-a} \\ t = \arctan \sqrt { \left( x-a \over b-x\right) } \]
we get pretty close to |dw:1349076028513:dw|
this is pretty close from solution of WA
Cleaning up the solution i get \[ \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right) \over 8 \]
Why should we express t in terms of arc tan?
I don't know ... I just want match up our answer with WA
Perhaps... That help get a prettier t... \[tan t = sint / cost = \sqrt{\frac{x-a}{b-a}}\] *Still working on that*
Let's graph our solution
something is missing
\[ \tan t = \sin t / \cos t = \sqrt{\frac{a-x}{x-b}} \]
:'( I got into trouble!!! \[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}}{b-a}(1-\frac{x-a}{b-a})]+C \]\[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}(b-x)}{(b-a)^2}]+C \]\[=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})]+\frac{1}{4}\sqrt{(x-a)(b-x)}(b-x)+C \]Blah... Doesn't seem right :'(
I almost got same result as mathematica except for this one \[ \left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \]
\[ =-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}}{b-a}(1-2\frac{x-a}{b-a})]+C \] you missed 2 here.
Oh!!! Let me redo it....
also change that value inside arctan it should be \[ \sqrt{x-a \over b-x}\]
Once again, it doesn't look great at all... \[-\frac{1}{4}(a-b)^2 [ tan^{-1} (\sqrt{\frac{x-a}{b-x}}) - \frac{\sqrt{(x-a)(b-x)}}{b-a}[1-2(\frac{x-a}{b-a})]] +C\]\[=-\frac{1}{4}(a-b)^2 [ tan^{-1} (\sqrt{\frac{x-a}{b-x}}) - \frac{\sqrt{(x-a)(b-x)}}{b-a}(\frac{a+b-2x}{b-a})] +C\]\[=-\frac{1}{4}(a-b)^2 tan^{-1} (\sqrt{\frac{x-a}{b-x}}) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C\]
so far so good every part matches solution from WA except \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) \]
I meant what wolf showed you...
yep!! http://www.wolframalpha.com/input/?i=Integrate+sqrt%28%28a-x%29%28x-b%29%29
cancel those terms at the front
\[ \frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}} \]
Cancelling those first terms we get \[ {1 \over 8}\left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)\]
it seems we are correct!! Wolf is pretty nasty!!
We need to prove that \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) = \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \] this is not correct ... however I'll show you this is equivalent in integrals!!
No... not really, the coefficient of arctan (simpler one) is -1/4, while the coefficient of arctan (wolf) is -1/8..
|dw:1349079549133:dw|
there is a formula that \( \arctan(x) + \arctan(1/x)= {\pi \over 2}\) |dw:1349079814250:dw|
hence we have \[ \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) = {\pi \over 2} - \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\] But this \( \pi \over 2 \) is constant and you can ignore it. Hence the result of the wolf follows.
But what about the coefficient??
\[ =-\frac{1}{4}(a-b)^2 \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C \\ -\frac{1}{4}(a-b)^2 {1\over 2 }\left( {\pi \over 2} - \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C\\ \frac{1}{8}(a-b)^2 \left( \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) \\ +C -{\pi \over 16 } (a-b)^2\\ \text{ Just let } C -{\pi \over 16 } (a-b)^2 = C_1 \text{ some constnat} \]
Looks like i missed some - somewhere.
Ehhhhh... I love my solution *v*
yeah .. no need to go though rigorous way as Wolf.
the above is just equally right!!
It's always good know to how to get the wolf solution. Thanks !!!
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.
|dw:1349081258153:dw|
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)
Eh?! three ways to do it?!
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.
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 :\