## experimentX Group Title Integration: one year ago one year ago

1. experimentX Group Title

the answer of the integration is ${1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t + {\sin (4t)\over 4}\right)$

2. experimentX Group Title

On the other hand, Mathematica expresses solution as $\frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}}$

3. experimentX Group Title

this is minus ${1 \over 4} \sqrt{b^2 -a^2}(b-a)\left( t - {\sin (4t)\over 4}\right)$

4. Callisto Group Title

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?

5. experimentX Group Title

6. experimentX Group Title

yeah i made error ... - ( 1 - cos ..) = -

7. experimentX Group Title

If we put $\sin (t) = \sqrt{x-a\over b-a} \\ \cos (t) = \sqrt{b-x\over b-a} \\ t = \arctan \sqrt { \left( x-a \over b-x\right) }$

8. experimentX Group Title

we get pretty close to |dw:1349076028513:dw|

9. experimentX Group Title

this is pretty close from solution of WA

10. experimentX Group Title

Cleaning up the solution i get $\left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right) \over 8$

11. Callisto Group Title

Why should we express t in terms of arc tan?

12. experimentX Group Title

I don't know ... I just want match up our answer with WA

13. Callisto Group Title

Perhaps... That help get a prettier t... $tan t = sint / cost = \sqrt{\frac{x-a}{b-a}}$ *Still working on that*

14. experimentX Group Title

Let's graph our solution

15. experimentX Group Title

something is missing

16. experimentX Group Title

$\tan t = \sin t / \cos t = \sqrt{\frac{a-x}{x-b}}$

17. Callisto Group Title

:'( 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 :'(

18. experimentX Group Title

I almost got same result as mathematica except for this one $\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]$

19. experimentX Group Title

$=-\frac{1}{4} (a-b)^2 [\tan^{-1} (\sqrt{\frac{x-a}{b-a}})-\frac{\sqrt{(x-a)(b-x)}}{b-a}(1-2\frac{x-a}{b-a})]+C$ you missed 2 here.

20. Callisto Group Title

Oh!!! Let me redo it....

21. experimentX Group Title

also change that value inside arctan it should be $\sqrt{x-a \over b-x}$

22. Callisto Group Title

Sorry.... typos!!

23. Callisto Group Title

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$

24. experimentX Group Title

so far so good every part matches solution from WA except $\tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right )$

25. Callisto Group Title

What is that in WA?

26. experimentX Group Title

Wolf ...

27. Callisto Group Title

I meant what wolf showed you...

28. experimentX Group Title
29. experimentX Group Title

cancel those terms at the front

30. experimentX Group Title

$\frac{\sqrt{a-x} \sqrt{-b+x} \left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)}{8 \sqrt{(a-x) (-b+x)}}$

31. experimentX Group Title

Cancelling those first terms we get ${1 \over 8}\left(2 \sqrt{a-x} \sqrt{-b+x} (-a-b+2 x)-(a-b)^2 \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right]\right)$

32. experimentX Group Title

it seems we are correct!! Wolf is pretty nasty!!

33. experimentX Group Title

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!!

34. Callisto Group Title

No... not really, the coefficient of arctan (simpler one) is -1/4, while the coefficient of arctan (wolf) is -1/8..

35. experimentX Group Title

|dw:1349079549133:dw|

36. experimentX Group Title

there is a formula that $$\arctan(x) + \arctan(1/x)= {\pi \over 2}$$ |dw:1349079814250:dw|

37. experimentX Group Title

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.

38. Callisto Group Title

39. experimentX Group Title

$=-\frac{1}{4}(a-b)^2 \tan^{-1} \left(\sqrt{\frac{x-a}{b-x}}\right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C \\ -\frac{1}{4}(a-b)^2 {1\over 2 }\left( {\pi \over 2} - \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) +C\\ \frac{1}{8}(a-b)^2 \left( \text{ArcTan}\left[\frac{a+b-2 x}{2 \sqrt{a-x} \sqrt{-b+x}}\right] \right ) + \frac{1}{4}\sqrt{(x-a)(b-x)}(a+b-2x) \\ +C -{\pi \over 16 } (a-b)^2\\ \text{ Just let } C -{\pi \over 16 } (a-b)^2 = C_1 \text{ some constnat}$

40. experimentX Group Title

Looks like i missed some - somewhere.

41. Callisto Group Title

Ehhhhh... I love my solution *v*

42. experimentX Group Title

yeah .. no need to go though rigorous way as Wolf.

43. experimentX Group Title

the above is just equally right!!

44. Callisto Group Title

It's always good know to how to get the wolf solution. Thanks !!!

45. experimentX Group Title

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.

46. Callisto Group Title

Why.......

47. experimentX Group Title

|dw:1349081258153:dw|

48. experimentX Group Title

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)

49. Callisto Group Title

Eh?! three ways to do it?!

50. experimentX Group Title

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.

51. Callisto Group Title

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 :\