Question

Integration:

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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?

Answer 5

looks like i made error

Answer 6

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

Answer 7

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

Answer 8

we get pretty close to