Answer:
Option A,C
Explanation:
Let $I_{1}=\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
=$\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
+ $\int_{\pi}^{2\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
+ $\int_{2\pi}^{3\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
+$\int_{3\pi}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
$\therefore$ $ I_{1}=I_{2}+I_{3}+I_{4}+I_{5}$ ..........(i)
Now, $I_{3}=\int_{\pi}^{2\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
Put $t=\pi+t\Rightarrow dt=dt$
$\therefore$ $I_{3}=\int_{0}^{\pi} e^{\pi+t}(\sin^{6}at+\cos^{6}at)dt$
= $e^{t}.I_{2}$ .........(ii)
Now,
I4= $\int_{2\pi}^{3\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
Put $t=2\pi+t\Rightarrow dt=dt$
$\therefore$ $I_{4}=\int_{0}^{\pi} e^{2\pi+t}(\sin^{6}at+\cos^{6}at)dt$
= $e^{2 \pi}.I_{2}$ ..........(iii)
and I5= $\int_{3\pi}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt$
Put $t=3\pi+t\Rightarrow dt=dt$
$\therefore$ $I_{5}=\int_{0}^{\pi} e^{3\pi+t}(\sin^{6}at+\cos^{6}at)dt$
= $e^{3 \pi}.I_{2}$ .........(iv)
From Eqs. (i), (ii) ,(iii) and (iv) , we get
$I_{1}=I_{2}+e^{\pi}I_{2}+e^{2\pi}I_{2}+e^{3\pi}I_{2}$
$=I_{2}(1++e^{\pi}+e^{2\pi}+e^{3\pi})$
$L= \frac{\int_{0}^{4\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt}{\int_{0}^{\pi} e^{t}(\sin^{6}at+\cos^{6}at)dt}$
= $(1+e^{\pi}+e^{2\pi}+e^{3\pi})$
= $\frac{1-(e^{4\pi}-1)}{e^{\pi}-1} $ for $a \epsilon R$