Question

∫√{（x-a）（b-x）} dx 求不定积分,注意不是∫[1/√（x-a）（b-x）dx 求思路或者过程

Answer 1

∫ √[(x-a)(b-x)] dx
The substitution x = acos²t + bsin²t
dx = 2(-acostsint + bsintcost)dt = 2(b-a)sintcost dt
(x-a) = (b-a)sin²t，(b-x) = (b-a)cos²t
Then J = ∫ √[(x-a)(b-a)] dx
= √[(b-a)²sin²tcos²t] * 2(b-a)sintcost dt
= 2(b-a)²∫ sin²tcos²t dt
= (1/2)(b-a)²∫ sin²(2t) dt
= (1/4)(b-a)²∫ (1-cos(4t)) dt
= (1/4)(b-a)²(t-(1/4)sin(4t)) + C
Using sint = √[(x-a)/(b-a)]，cost = √[(b-x)/(b-a)]
sin(4t) = 4 * √[(x-a)/(b-a)] * √[(b-x)/(b-a)] * [(b-x)/(b-a) - (x-a)/(b-a)]
J = (1/4)(b-a)² * {arcsin√[(x-a)/(b-a)] - √[(x-a)(b-x)] / (b-a)² * (a+b-2x)} + C

Alternatives：
(x-a)(b-x) = (x-a)b - (x-a)x = bx - ab - x² + ax = -x² + (a+b)x - ab
Let -x² + (a+b)x - ab = -(x+B)²+C = -(x² + 2Bx + B²)+C = -x² - 2Bx + (C-B²)
By comparing the coefficients of both sides，we have the following equations：
a+b = -2B => B = (-1/2)(a+b)
-ab = C-B² => C = B²-ab = (1/4)(a+b)²-ab = (a-b)²/4
Therefore = (x-a)(b-x) = -[x - (a+b)/2]² + (a-b)²/4

J = ∫ √[(x-a)(b-x)] dx = ∫ √{(a-b)²/4 - [x - (a+b)/2]²} dx
Let x - (a+b)/2 = (a-b)/2 * sinθ，dx = (a-b)/2 * cosθdθ

J = (a-b)/2 * ∫ √{(a-b)²/4 - [√[(a-b)²/4] * sinθ]²} cosθdθ
= (a-b)/2 * ∫ √[(a-b)²/4 * cos²θ] cosθdθ
= (a-b)²/4 * ∫ cos²θ dθ
= (a-b)²/8 * ∫ (1+cos2θ) dθ
= (a-b)²/8 * [θ + (sin2θ)/2] + C
= (a-b)²/8 * (θ + sinθcosθ) + C

x - (a+b)/2 = (a-b)/2 * sinθ
sinθ = [x - (a+b)/2] / [(a-b)/2] = (-a-b+2x)/(a-b)
cosθ = √(1-sin²θ) = 2√[(x-a)(b-x)]/(a-b)
真是复杂，代回就可以了。

Answer 2

∫√{（x-a）（b-x）} dx=∫√{-x^2+(a+b)x-ab}dx
=∫√{-(x-(a+b/2))^2+((a+b)/2)^2-ab}dx
=∫√{((a-b)/2)^2-(x-(a+b)/2)^2}dx

变为∫√{R^2-(x-A)^2}dx的形式 令x=A+Rsint 代入求解
∫Rcost *（R*cost）dt
=∫R^2*(cos2t+1）/2 dt
=R^2(t+*sin2t)/2+C
=R^2(arcsin((x-A)/R)-sin(2*(x-A)/R)))/2+C
R=(a-b)/2 A=(a+b)/2