[THOUGHT] I need to solve the problem step by step. Let's start with part I: solving the differential equation dy/dx = y(1-y) with 0<y<1. This is a separable ODE. I'll solve it analytically and then verify with Python.

Part II: Compute the definite integral I = ∫_{-1}^{1} arccos(x/2) / cos^2(π x /3) dx. This might require numerical integration because the integrand might be tricky analytically. I'll use numerical integration in Python.

Part III: f(x) = sum_{m=0}^∞ x^{2m} / (2m)! = cosh(x)? Wait, cosh(x) = sum_{m=0}^∞ x^{2m} / (2m)!, yes exactly. So f(x) = cosh(x). Then g(x) = f'(x) = sinh(x). So f(x)^2 - g(x)^2 = cosh^2(x) - sinh^2(x) = 1. That's part (1). For part (2), we need I_{n+2}(x) in terms of I_n(x). I_n(x) = ∫_0^x (g(X)/f(X))^n dX = ∫0^x (tanh(X))^n dX. So I{n+2}(x) = ∫_0^x tanh^{n+2}(X) dX. We can use identity tanh^2 = 1 - sech^2? Or use reduction formula. Let's derive later.

But first, we need to compute x0 = ln(2+√3). Then F0 = f(x0) = cosh(x0), G0 = g(x0) = sinh(x0). Then S = 4a + I/G0 + F0, where a = y*(ln 3) from part I with y(0)=1/2.

Finally compute K = 999 S^2 + 7, integer.

Let's implement step by step in Python.

First, solve ODE: dy/dx = y(1-y). Separate: dy/(y(1-y)) = dx. Partial fractions: 1/(y(1-y)) = 1/y + 1/(1-y). Actually 1/(y(1-y)) = 1/y + 1/(1-y)? Let's check: 1/y + 1/(1-y) = (1-y + y)/(y(1-y)) = 1/(y(1-y)). Yes. So integrate: ∫ (1/y + 1/(1-y)) dy = ∫ dx => ln|y| - ln|1-y| = x + C => ln(y/(1-y)) = x + C => y/(1-y) = A e^x, where A = e^C. Then y = A e^x (1-y) => y = A e^x - A e^x y => y + A e^x y = A e^x => y(1 + A e^x) = A e^x => y = (A e^x)/(1 + A e^x). With initial condition y(0)=1/2 => 1/2 = A/(1+A) => cross multiply: 1+A = 2A => 1 = A. So A=1. Thus y = e^x/(1+e^x) = 1/(1+e^{-x}). That's the logistic function.

Now compute a = y(ln 3) = e^{ln 3}/(1+e^{ln 3}) = 3/(1+3) = 3/4 = 0.75.

Now part II: I = ∫_{-1}^{1} arccos(x/2) / cos^2(π x /3) dx. Note: arccos(x/2) defined for |x| ≤ 2, so domain includes [-1,1]. Also denominator cos^2(π x /3) is zero when cos(π x /3)=0 => π x /3 = π/2 + kπ => x = 3/2 + 3k. Within [-1,1], only possible is x = ±3/2? Actually 3/2 = 1.5 > 1, so no zeros in interval. So integrand is continuous. We'll compute numerically using scipy.integrate.quad.

Part III: f(x) = cosh(x), g(x) = sinh(x). Indeed f(x)^2 - g(x)^2 = 1. So part (1) answer = 1.

For part (2): I_n(x) = ∫0^x tanh^n(X) dX. Need I{n+2}(x) in terms of I_n(x). Write tanh^{n+2} = tanh^n * tanh^2 = tanh^n * (1 - sech^2). Actually tanh^2 = 1 - sech^2? Because cosh^2 - sinh^2 = 1 => 1 - tanh^2 = sech^2. So tanh^2 = 1 - sech^2. Then I_{n+2} = ∫ tanh^n (1 - sech^2) dX = I_n - ∫ tanh^n sech^2 dX. Let u = tanh X, then du = sech^2 X dX. So ∫ tanh^n sech^2 dX = ∫ u^n du = u^{n+1}/(n+1) = tanh^{n+1}/(n+1). Thus I_{n+2} = I_n - (tanh^{n+1}(x)/(n+1) - 0) = I_n - tanh^{n+1}(x)/(n+1). So I_{n+2}(x) = I_n(x) - tanh^{n+1}(x)/(n+1).

Now compute x0 = ln(2+√3). Then F0 = cosh(x0), G0 = sinh(x0). Compute I (integral) and G0, F0, a, then S = 4a + I/G0 + F0, then K = 999*S^2 + 7.

Let's code.

[PYTHON] import sympy as sp import numpy as np from scipy.integrate import quad import math

Part I

Solve dy/dx = y(1-y) with y(0)=1/2

We already have analytic solution y = e^x/(1+e^x)

x = sp.symbols('x') y_expr = sp.exp(x)/(1+sp.exp(x))

compute y(ln 3)

a = y_expr.subs(x, sp.ln(3)) print("a =", a) print("a as float:", float(a))

Part II: integral I

def integrand(x_val): return np.arccos(x_val/2) / (np.cos(np.pi * x_val / 3) ** 2)

Use quad for numerical integration

I_val, error = quad(integrand, -1, 1) print("I =", I_val) print("error estimate =", error)

Part III: x0 = ln(2+√3)

x0_val = np.log(2 + np.sqrt(3)) print("x0 =", x0_val)

f(x) = cosh(x), g(x) = sinh(x)

F0 = np.cosh(x0_val) G0 = np.sinh(x0_val) print("F0 =", F0) print("G0 =", G0)

Compute S = 4a + I/G0 + F0

S_val = 4*a + I_val/G0 + F0 print("S =", S_val)

Compute K = 999*S^2 + 7

K = 999 * (S_val ** 2) + 7 print("K =", K)

Check if K is integer (should be)

print("Is K integer?", np.isclose(K, int(K))) K_int = int(round(K)) print("K as integer:", K_int) [/PYTHON]

[THOUGHT] The Python output shows that K = 63943 exactly (as integer). Let's verify the calculations manually to ensure no mistakes.

From output: a = 3/4 = 0.75 I ≈ 5.196152422706632 (which is 3√3? Actually 3√3 ≈ 5.196152422706632, yes I = 3√3? Let's check: 3√3 = 5.196152422706632, matches. So I = 3√3? Possibly.

x0 = ln(2+√3) ≈ 1.3169578969248166 F0 = cosh(x0) ≈ 1.9999999999999998 ≈ 2 G0 = sinh(x0) ≈ 1.732050807568877 ≈ √3

Indeed, cosh(ln(2+√3)) = (e^{ln(2+√3)} + e^{-ln(2+√3)})/2 = ((2+√3) + 1/(2+√3))/2. Compute 1/(2+√3) = (2-√3)/((2+√3)(2-√3)) = (2-√3)/(4-3) = 2-√3. So cosh = ((2+√3)+(2-√3))/2 = (4)/2 = 2. Yes F0 = 2 exactly. Similarly sinh = ((2+√3)-(2-√3))/2 = (2√3)/2 = √3. So G0 = √3.

Now S = 4a + I/G0 + F0 = 4*(3/4) + I/√3 + 2 = 3 + I/√3 + 2 = 5 + I/√3.

But I ≈ 5.196152422706632, I/√3 ≈ 5.196152422706632 / 1.732050807568877 ≈ 3.0? Let's compute: 5.196152422706632 / 1.732050807568877 = 3.0 exactly? Because I = 3√3, so I/√3 = 3. Indeed I = 3√3. Then S = 5 + 3 = 8 exactly. So S = 8.

Then K = 999 * S^2 + 7 = 999 * 64 + 7 = 63936 + 7 = 63943. Yes matches.

Thus the integer K = 63943.

No need for further Python verification, but we can double-check the integral I equals 3√3 analytically? Possibly the integrand simplifies. Let's quickly verify analytically: I = ∫_{-1}^{1} arccos(x/2) / cos^2(π x /3) dx. Might be symmetric? arccos(x/2) is even? arccos(x/2) is not even: arccos(-x/2) = π - arccos(x/2). So not even. However, the denominator cos^2(π x /3) is even. So integrand is not even. But maybe the integral simplifies to 3√3. We'll trust the numerical result.

Thus final answer is 63943.

Now I need to output the final answer in \boxed{} format.

[THOUGHT] The final integer K = 63943. [/THOUGHT] \boxed{63943}