Evaluating Ratio of Exponential Trigonometric Definite Integrals

To find the values of $a$ and $L$ that satisfy the given relation, let us define the function: $f(t) = \sin^6(at) + \cos^4(at)$

First, we analyze the periodicity of $f(t)$ for an integer value of $a$ (such as $a = 2$ or $a = 4$).

Using the trigonometric addition formulas for any $a \in \mathbb{Z}$: $\sin(a(t + \pi)) = \sin(at + a\pi) = \sin(at)\cos(a\pi) + \cos(at)\sin(a\pi) = (-1)^a \sin(at)$ $\cos(a(t + \pi)) = \cos(at + a\pi) = \cos(at)\cos(a\pi) - \sin(at)\sin(a\pi) = (-1)^a \cos(at)$

Since the powers of $\sin(at)$ and $\cos(at)$ in $f(t)$ are even (6 and 4 respectively): $\sin^6(a(t + \pi)) = \left[(-1)^a \sin(at)\right]^6 = (-1)^{6a} \sin^6(at) = \sin^6(at)$ $\cos^4(a(t + \pi)) = \left[(-1)^a \cos(at)\right]^4 = (-1)^{4a} \cos^4(at) = \cos^4(at)$

Thus, for any integer $a$: $f(t + \pi) = f(t)$ This implies that $f(t)$ is periodic with a period of $\pi$ (or a fraction thereof), so $f(t + k\pi) = f(t)$ for any integer $k$.

Now, let $I_1$ be the numerator integral: $I_1 = \int_{0}^{4\pi} e^t f(t) \, dt$

We can split the interval $[0, 4\pi]$ into four subintervals of length $\pi$: $I_1 = \sum_{k=0}^{3} \int_{k\pi}^{(k+1)\pi} e^t f(t) \, dt$

For each integral in the summation, substitute $t = x + k\pi$, which gives $dt = dx$: $\int_{k\pi}^{(k+1)\pi} e^t f(t) \, dt = \int_{0}^{\pi} e^{x + k\pi} f(x + k\pi) \, dx$

Using the periodicity property $f(x + k\pi) = f(x)$: $\int_{0}^{\pi} e^{x + k\pi} f(x) \, dx = e^{k\pi} \int_{0}^{\pi} e^x f(x) \, dx$

Therefore, $I_1$ becomes: $I_1 = \left( \sum_{k=0}^{3} e^{k\pi} \right) \int_{0}^{\pi} e^x f(x) \, dx = \left( 1 + e^{\pi} + e^{2\pi} + e^{3\pi} \right) \int_{0}^{\pi} e^t f(t) \, dt$

Let $I_2$ be the denominator integral: $I_2 = \int_{0}^{\pi} e^t f(t) \, dt$

The ratio $L = \frac{I_1}{I_2}$ is: $L = \frac{\left( 1 + e^{\pi} + e^{2\pi} + e^{3\pi} \right) \int_{0}^{\pi} e^t f(t) \, dt}{\int_{0}^{\pi} e^t f(t) \, dt} = 1 + e^{\pi} + e^{2\pi} + e^{3\pi}$

Summing this finite geometric series with common ratio $r = e^{\pi}$ and $N = 4$ terms: $L = \frac{(e^{\pi})^4 - 1}{e^{\pi} - 1} = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$

Since this property holds for any integer $a$, it holds specifically for both $a = 2$ and $a = 4$:

• For $a = 2$, $L = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$
• For $a = 4$, $L = \frac{e^{4\pi} - 1}{e^{\pi} - 1}$

Therefore, the correct options are A and C.