Let's do some calculus! (8)

Let $N = \int\limits_{0}^{4\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt$

$D = \int\limits_{0}^{\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt$

We have to find $\dfrac{N}{D}$ .

$N = \int\limits_{0}^{4\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt$

$\text{ } = \int\limits_{0}^{\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{\pi}^{2\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{2\pi}^{3\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{3\pi}^{4\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt$

After the following subsitutions: $t-\pi=u, t-2\pi=v, t-3\pi=w$ in the second, third and fourth integrals respectively, it is seen that the trigonometry functions do not change regardless of $a$ being $2$ or $4$ . Since it is definite integration, $t$ is a dummy variable and therefore:

$N = \int\limits_{0}^{\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{0}^{\pi}{e^{t+\pi}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{0}^{\pi}{e^{t+2\pi}\left(\sin^{6}at + \cos^{4}at\right)} \,dt + \int\limits_{0}^{\pi}{e^{t+3\pi}\left(\sin^{6}at + \cos^{4}at\right)} \,dt$

$N = (1 + e^\pi + e^{2\pi} + e^{3\pi})\int\limits_{0}^{\pi}{e^{t}\left(\sin^{6}at + \cos^{4}at\right)} \,dt = (1 + e^\pi + e^{2\pi} + e^{3\pi})\cdot D$

$\implies \dfrac{N}{D} = (1 + e^\pi + e^{2\pi} + e^{3\pi}) = \boxed{ \dfrac{e^{4\pi}-1}{e^\pi - 1} }$

(The last equation was simplified using the formula for geometric progression.)