The option(s) with the values of $a$ and $L$ that satisfy the following equation is(are) $\dfrac{\int_{0}^{4\pi} e^{t}\left(\sin^{6} at + \cos^{4} at\right) dt}{\int_{0}^{\pi} e^{t}\left(\sin^{6} at + \cos^{4} at\right) dt} = L?$

$N = \int_0^{4\pi} e^t\left(\sin^6 at+\cos^4 at\right)dt$

$D = \int_0^{\pi} e^t\left(\sin^6 at+\cos^4 at\right)dt$

$\dfrac{N}{D} = L$

$\text{To find: } a \text{ and } L \text{ (among the options)}$

$f(t)=\sin^6 at+\cos^4 at$ $a\in\mathbb{Z}:\ \sin(a(t+\pi))=\pm\sin(at),\ \text{and powers are even}$ $\Rightarrow f(t+\pi)=f(t)$ $1.\ \text{split } N \text{ into } 4 \text{ pieces of width } \pi$ $2.\ \text{shift } t=u+k\pi,\ \text{use } f(t+\pi)=f(t)$ $3.\ L = \text{a geometric sum}$ $N = \sum_{k=0}^{3}\int_{k\pi}^{(k+1)\pi} e^t f(t)\,dt$ $\int_{k\pi}^{(k+1)\pi} e^t f(t)\,dt = e^{k\pi}\int_0^{\pi} e^u f(u)\,du = e^{k\pi} D$ $N = \left(\sum_{k=0}^{3}e^{k\pi}\right)D = \left(1+e^{\pi}+e^{2\pi}+e^{3\pi}\right)D$ $L = 1+e^{\pi}+e^{2\pi}+e^{3\pi} = \dfrac{e^{4\pi}-1}{e^{\pi}-1}$ $L \text{ independent of } a \ (\text{integer } a)$ $\text{(1) } a=2\ \checkmark,\quad \text{(3) } a=4\ \checkmark;\quad \text{(2),(4) wrong}$ $\boxed{\text{options (1) and (3)}}$