[THOUGHT] I will:

1. Symbolically solve the ODE in I and get a, then 4a.
2. Symbolically compute the integral in II.
3. Recognize f and g in III as cosh and sinh, verify f^2 - g^2, and derive the recursion, though for K only f and g at x0 are needed.
4. Use sympy to compute F0 = f(x0) = cosh(x0), G0 = g(x0) = sinh(x0) for x0 = ln(2+√3).
5. Form S = 4a + I/G0 + F0 and then K = 999*S^2 + 7, simplify exactly to an integer. [/THOUGHT] [PYTHON] import sympy as sp

Part I: solve dy/dx = y(1-y), find y*(x) with y(0)=1/2, then a = y*(ln 3)

x = sp.symbols('x') y = sp.Function('y') ode = sp.Eq(sp.diff(y(x), x), y(x)*(1 - y(x))) solution_general = sp.dsolve(ode) solution_general

Impose y(0) = 1/2

C1 = sp.symbols('C1') y_expr = solution_general.rhs C1_solution = sp.solve(sp.Eq(y_expr.subs(x, 0), sp.Rational(1,2)), C1)[0] y_star = sp.simplify(y_expr.subs(C1, C1_solution)) a = sp.simplify(y_star.subs(x, sp.log(3))) four_a = sp.simplify(4a) print("y(x) =", y_star) print("a =", a) print("4a =", four_a)

Part II: compute I = ∫_{-1}^1 arccos(x/2)/cos^2(pi x/3) dx

x = sp.symbols('x', real=True) I = sp.integrate(sp.acos(x/2)/sp.cos(sp.pi*x/3)**2, (x, -1, 1)) I_simplified = sp.simplify(I) print("Integral I =", I_simplified)

Part III: define f and g, verify f^2 - g^2, and note recursion (symbolic check)

x = sp.symbols('x', real=True, positive=True) m = sp.symbols('m', integer=True, nonnegative=True)

Define f via series sum

f_series = sp.summation(x**(2m)/sp.factorial(2m), (m, 0, sp.oo)) g_series = sp.diff(f_series, x) f_simplified = sp.simplify(f_series) g_simplified = sp.simplify(g_series) fg_diff = sp.simplify(f_simplified2 - g_simplified2) print("f(x) =", f_simplified) print("g(x) =", g_simplified) print("f(x)^2 - g(x)^2 =", fg_diff)

Part III recursion: confirm tanh representation

t = sp.symbols('t', real=True) n = sp.symbols('n', integer=True, nonnegative=True) lhs = sp.tanh(t)**(n+2) rhs = sp.tanh(t)n - (1/(n+1))*sp.diff(sp.tanh(t)(n+1), t) recursion_check = sp.simplify(lhs - rhs) print("Recursion identity check (should be 0):", recursion_check)

Part IV: compute x0, F0, G0, S, and K

x0 = sp.log(2 + sp.sqrt(3)) F0 = sp.cosh(x0) G0 = sp.sinh(x0) F0_simplified = sp.simplify(F0) G0_simplified = sp.simplify(G0) print("x0 =", x0) print("F0 =", F0_simplified) print("G0 =", G0_simplified)

S = sp.simplify(four_a + I_simplified/G0_simplified + F0_simplified) K = sp.simplify(999*S**2 + 7) print("S =", S) print("K =", K) [/PYTHON]

[THOUGHT] From the Python output we had:

• (y_(x) = \dfrac{e^x}{e^x+1}), hence (a = y_(\ln 3) = \dfrac{3}{4}), so (4a = 3).
• (f(x) = \cosh x,\ g(x) = \sinh x).
• With (x_0 = \ln(2+\sqrt{3})), we get [ F_0 = \cosh x_0 = 2,\quad G_0 = \sinh x_0 = \sqrt{3}. ]

To compute [ I = \int_{-1}^{1}\frac{\arccos(x/2)}{\cos^2(\pi x/3)},dx, ] we integrate by parts with (u = \arccos(x/2)), (dv = \sec^2(\pi x/3),dx). Then [ du = -\frac{dx}{\sqrt{4 - x^2}},\quad v = \frac{3}{\pi}\tan\left(\frac{\pi x}{3}\right). ] So [ I = \left.\frac{3}{\pi}\arccos\left(\frac{x}{2}\right)\tan\left(\frac{\pi x}{3}\right)\right|{-1}^{1} +\frac{3}{\pi}\int{-1}^{1} \frac{\tan(\pi x/3)}{\sqrt{4-x^2}},dx. ] The boundary term evaluates to [ \frac{3}{\pi}\Big[\arccos(1/2)\tan(\pi/3) - \arccos(-1/2)\tan(-\pi/3)\Big] = \sqrt{3} - (-2\sqrt{3}) = 3\sqrt{3}. ] For the remaining integral, set (x = 2\sin\theta), so (\sqrt{4-x^2}=2\cos\theta), (dx = 2\cos\theta,d\theta), and the limits become (\theta\in[-\pi/6,\pi/6]). Then [ \int_{-1}^{1}\frac{\tan(\pi x/3)}{\sqrt{4-x^2}},dx = \int_{-\pi/6}^{\pi/6}\tan\Big(\tfrac{2\pi}{3}\sin\theta\Big),d\theta. ] The integrand is odd in (\theta), so the integral is (0). Thus [ I = 3\sqrt{3}. ]

Now [ S = 4a + \frac{I}{G_0} + F_0 = 3 + \frac{3\sqrt{3}}{\sqrt{3}} + 2 = 3 + 3 + 2 = 8. ] Finally, [ \mathcal{K} = 999S^2 + 7 = 999\cdot 64 + 7 = 63943. ] [/THOUGHT] \boxed{63943}