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) $a=4, L=\frac{e^{4 \pi}+1}{e^\pi+1}$
Answer & explanation
Correct answer: option 1
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