Step-by-step Solution

1. Analyze the structure of the given equation. The equation is given by: $L = \frac{\int_0^{4\pi} e^t (\sin^6(at) + \cos^4(at)) dt}{\int_0^\pi e^t (\sin^6(at) + \cos^4(at)) dt}$ Let's define the function $f(t) = \sin^6(at) + \cos^4(at)$. The equation becomes: $L = \frac{\int_0^{4\pi} e^t f(t) dt}{\int_0^\pi e^t f(t) dt}$

2. Determine the period of the function $f(t)$. The function $f(t)$ is a sum of two trigonometric functions.

   • The period of $\sin^k(ax)$ is $\frac{\pi}{|a|}$ if $k$ is an even integer.
   • The period of $\cos^k(ax)$ is $\frac{\pi}{|a|}$ if $k$ is an even integer. In our case, the powers are 6 and 4, which are even. So, the period of $\sin^6(at)$ is $\frac{\pi}{a}$ and the period of $\cos^4(at)$ is $\frac{\pi}{a}$ (assuming $a>0$ as is typical). The period of the sum of two functions with the same period $T$ is also $T$ (or a submultiple, but let's check). Let $T = \frac{\pi}{a}$. $f(t+T) = \sin^6(a(t+\pi/a)) + \cos^4(a(t+\pi/a)) = \sin^6(at+\pi) + \cos^4(at+\pi)$ $= (-\sin(at))^6 + (-\cos(at))^4 = \sin^6(at) + \cos^4(at) = f(t)$ To confirm this is the fundamental period, we can check $T/2 = \pi/(2a)$: $f(t+\frac{\pi}{2a}) = \sin^6(at+\frac{\pi}{2}) + \cos^4(at+\frac{\pi}{2}) = \cos^6(at) + \sin^4(at)$. Setting $f(t) = f(t+\frac{\pi}{2a})$ implies $\sin^6(at) + \cos^4(at) = \cos^6(at) + \sin^4(at)$, which is not true for all $t$. Thus, the fundamental period of $f(t)$ is $T = \frac{\pi}{a}$.

3. Use the property of integrals involving periodic functions. There is a useful property for integrals of the form $\int e^{kx} g(x) dx$ where $g(x)$ is periodic. For $k=1$, the property is: If $f(t)$ is a periodic function with period $T$, then for any integer $n$, $\int_0^{nT} e^t f(t) dt = \left( \frac{e^{nT} - 1}{e^T - 1} \right) \int_0^T e^t f(t) dt$ This property holds provided the upper limit of integration is an integer multiple of the period $T$.

4. Apply the property to the numerator and denominator. Let's check if the integration limits are integer multiples of the period $T = \pi/a$ for the given options where $a$ is an integer.

   • Numerator: The upper limit is $4\pi$. We can write $4\pi = nT = n(\pi/a)$, which implies $n = 4a$. If $a$ is an integer, $n$ is also an integer.
   • Denominator: The upper limit is $\pi$. We can write $\pi = mT = m(\pi/a)$, which implies $m = a$. If $a$ is an integer, $m$ is also an integer.

   Since the options provide integer values for $a$ ($a=2, a=4$), we can apply the formula. Let $I_{num} = \int_0^{4\pi} e^t f(t) dt$ and $I_{den} = \int_0^\pi e^t f(t) dt$. With $n=4a$ and $m=a$: $I_{num} = \int_0^{(4a)T} e^t f(t) dt = \left( \frac{e^{(4a)T} - 1}{e^T - 1} \right) \int_0^T e^t f(t) dt = \left( \frac{e^{4\pi} - 1}{e^{\pi/a} - 1} \right) \int_0^{\pi/a} e^t f(t) dt$ $I_{den} = \int_0^{aT} e^t f(t) dt = \left( \frac{e^{aT} - 1}{e^T - 1} \right) \int_0^T e^t f(t) dt = \left( \frac{e^{\pi} - 1}{e^{\pi/a} - 1} \right) \int_0^{\pi/a} e^t f(t) dt$

5. Calculate the value of L. $L = \frac{I_{num}}{I_{den}} = \frac{\left( \frac{e^{4\pi} - 1}{e^{\pi/a} - 1} \right) \int_0^{\pi/a} e^t f(t) dt}{\left( \frac{e^{\pi} - 1}{e^{\pi/a} - 1} \right) \int_0^{\pi/a} e^t f(t) dt}$ The integral $\int_0^{\pi/a} e^t f(t) dt$ is non-zero because the integrand $e^t(\sin^6(at) + \cos^4(at))$ is strictly positive for $t \in (0, \pi/a)$. Thus, we can cancel the integral terms. $L = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$ This result for $L$ is independent of the integer value of $a$.

6. Evaluate the given options.

   • Option A: $a = 2, L = \frac{e^{4\pi} - 1}{e^\pi - 1}$. Since $a=2$ is an integer, our derivation is valid. The calculated value of $L$ matches the one given in the option. So, Option A is correct.
   • Option B: $a = 2, L = \frac{e^{4\pi} + 1}{e^\pi + 1}$. The value of $L$ does not match our derived result. So, Option B is incorrect.
   • Option C: $a = 4, L = \frac{e^{4\pi} - 1}{e^\pi - 1}$. Since $a=4$ is an integer, our derivation is valid. The calculated value of $L$ matches the one given in the option. So, Option C is correct.
   • Option D: $a = 4, L = \frac{e^{4\pi} + 1}{e^\pi + 1}$. The value of $L$ does not match our derived result. So, Option D is incorrect.

Conclusion

Both options A and C satisfy the given equation.