## experimentX 3 years ago Integration:

1. experimentX

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

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

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

4. Callisto

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

6. experimentX

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

7. experimentX

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

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

9. experimentX

this is pretty close from solution of WA

10. experimentX

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

Why should we express t in terms of arc tan?

12. experimentX

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

13. Callisto

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

14. experimentX

Let's graph our solution

15. experimentX

something is missing

16. experimentX

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

17. Callisto

:'( 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

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

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

20. Callisto

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

21. experimentX

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

22. Callisto

Sorry.... typos!!

23. Callisto

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

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

25. Callisto

What is that in WA?

26. experimentX

Wolf ...

27. Callisto

I meant what wolf showed you...

28. experimentX
29. experimentX

cancel those terms at the front

30. experimentX

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

31. experimentX

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

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

33. experimentX

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

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

35. experimentX

|dw:1349079549133:dw|

36. experimentX

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

37. experimentX

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

39. experimentX

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

40. experimentX

Looks like i missed some - somewhere.

41. Callisto

Ehhhhh... I love my solution *v*

42. experimentX

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

43. experimentX

the above is just equally right!!

44. Callisto

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

45. experimentX

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

Why.......

47. experimentX

|dw:1349081258153:dw|

48. experimentX

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

Eh?! three ways to do it?!

50. experimentX

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

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 :\