The options(s) with the values of a and L that satisfy the following equation is (are) $\frac{\int\limits_0^{4 \pi} e^t\left(\sin ^6 a t+\cos ^4 a t\right) d t}{\int\limits_0^\pi e^t\left(\sin ^6 a t+\cos ^4 a t\right) d t}=L$?

(a) $a=2, L=\frac{e^{4 \pi}-1}{e^\pi-1}$
(b) $a=2, L=\frac{e^{4 \pi}+1}{e^\pi+1}$
(c) $a=4, L=\frac{e^{4 \pi}-1}{e^\pi-1}$
(d) $a=4, L=\frac{e^{4 \pi}+1}{e^\pi+1}$

(a), (c)

Let $f(t)=e^t\left(\sin ^6 a t+\cos ^4 a t\right)$. Then

$f(k \pi+t) =e^{k \pi+t}\left\{\sin ^6(a k \pi+a t)+\cos ^4(a k \pi+a t)\right\}$

$=e^{k \pi} e^t\left(\sin ^6 a t+\cos ^4 a t\right)$ for even value of a

$=e^{k \pi} f(t)$

∴   $\int\limits_0^{4 \pi} e^t\left(\sin ^6 a t+\cos ^4 a t\right) d t$

$=\int\limits_0^{4 \pi} f(t) d t=\sum\limits_{r=1}^4 \int\limits_{(r-1) \pi}^{r \pi} f(t) d t$

$=\sum\limits_{r=1}^4 \int\limits_0^\pi f((r-1) \pi+u) d u$, where $t=(r-1) \pi+u$

$=\sum\limits_{r=1}^4 \int\limits_0^\pi e^{(r-1) \pi} f(u) d u$

$=\sum\limits_{r=1}^4 e^{(r-1) \pi} \int\limits_0^\pi f(u) d u$

$=\left(1+e^\pi+e^{2 \pi}+e^{3 \pi}\right) \int\limits_0^\pi f(u) d u$

$=\left(\frac{e^{4 \pi}-1}{e^\pi-1}\right) \int\limits_0^\pi f(t) d t$

$\Rightarrow \frac{\int\limits_0^{4 \pi} e^t\left(\sin ^6 a t+\cos ^4 a t\right) d t}{\int\limits_0^\pi e^t\left(\sin ^6 a t+\cos ^4 a t\right) d t}=\frac{e^{4 \pi}-1}{e^\pi-1}$ for a = 2, 4