Definite Integration Question 50
Question 50
- 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) $a=4, L=\frac{e^{4 \pi}+1}{e^{\pi}+1}$
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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
$$ \begin{aligned} & I_{1}=I_{2}+e^{\pi} \cdot I_{2}+e^{2 \pi} \cdot I_{2}+e^{3 \pi} \cdot I_{2}=\left(1+e^{\pi}+e^{2 \pi}+e^{3 \pi}\right) I_{2} \ & \therefore \quad L=\frac{\int_{0}^{4 \pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t}{\int_{0}^{\pi} e^{t}\left(\sin ^{6} a t+\cos ^{4} a t\right) d t} \ & =\left(1+e^{\pi}+e^{2 \pi}+e^{3 \pi}\right) \ & =\frac{1 \cdot\left(e^{4 \pi}-1\right)}{e^{\pi}-1} \text { for } a \in R \end{aligned} $$
