Definite Integration Ques 48
- The option(s) with the values of $a$ and $L$ that satisfy the equation $\frac{\int _0^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t}{\int _0^{\pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t}=L$, is/are
(2015 Adv.)
(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) $a=4, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$
Show Answer
Solution:
- Let $I _1=\int _0^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t$
$$ =\int _0^{\pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t $$
$$ \begin{aligned} & +\int _{\pi}^{2 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t \\ & \quad+\int _{2 \pi}^{3 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t \\ & \quad+\int _{3 \pi}^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t \end{aligned} $$
$\therefore \quad I _1=I _2+I _3+I _4+I _5$
Now, $\quad I _3=\int _{\pi}^{2 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t$
Put $\quad t=\pi+x \Rightarrow \quad d t=d x$
$\therefore \quad I _3=\int _0^{\pi} e^{\pi+x} \cdot\left(\sin ^{6} a t+\cos ^{4} a t\right) d t=e^{\pi} \cdot I _2 \ldots$ (ii)
Now, $\quad I _4=\int _{2 \pi}^{3 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t$
Put $\quad t=2 \pi+x \Rightarrow d t=d x$
$\therefore \quad I _4=\int _0^{\pi} e^{x+2 \pi}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t=e^{2 \pi} \cdot I _2$
and $I _5=\int _{3 \pi}^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t$
Put $t=3 \pi+x$
$\therefore \quad I _5=\int _0^{\pi} e^{3 \pi+x}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t=e^{3 \pi} \cdot I _2$ From Eqs. (i), (ii), (iii) and (iv), we get
$$ \begin{aligned} & I _1=I _2+e^{\pi} \cdot I _2+e^{2 \pi} \cdot I _2+e^{3 \pi} \cdot I _2=\left(1+e^{\pi}+e^{2 \pi}+e^{3 \pi}\right) I _2 \\ & \therefore \quad L=\frac{\int _0^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t}{\int _0^{\pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t} \\ & =\left(1+e^{\pi}+e^{2 \pi}+e^{3 \pi}\right) \\ & =\frac{1 \cdot\left(e^{4 \pi}-1\right)}{e^{\pi}-1} \text { for } a \in R \end{aligned} $$
BETA
