Question

The option with the values of L that satisfying the equation $\frac{{\int }_0^{4\pi }{e}^t(si{n}^{6}at+co{s}^{4}at)dt}{{\int }_0^{\pi }{e}^t(si{n}^{6}at+co{s}^{4}at)dt}=L,$ is

• A. $L=\frac{e^{4\pi }-1}{e^{\pi }-1}$
  
• B. $L=\frac{e^{4\pi }+1}{e^{\pi }}+1$
  
• C. $L=\frac{e^{4\pi }-1}{e^{\pi }}-1$
  
• D. $L=\frac{e^{4\pi }+1}{e^{\pi }}-1$

Answer 1

The correct option is A

$L=\frac{e^{4\pi }-1}{e^{\pi }-1}$

$I_1={\int }_0^{4\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt={\int }_0^{\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt+{\int }_{\pi }^{2\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt+{\int }_{2\pi }^{3\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt+{}^{}{\int }_{3\pi }^{4\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt\therefore I_1=I_2+I_3+I_4+I_{5}Now,I_3={\int }_{\pi }^{2\pi }(e^{t}si{n}^{6}at+co{s}^{6}at)dtPutt=\pi +t\Rightarrow dt=dt\therefore I_3={\int }_0^{\pi }{e}^{\pi +t}.(si{n}^{6}at+co{s}^{6}at)dt=e^{\pi }.I_2...\left(ii\right)Now,I_4={\int }_{2\pi }^{3\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt=e^{2\pi }.I_2...\left(iii\right)$
$and{I}_5={\int }_{3\pi }^{4\pi }{e}^t(si{n}^{6}at+co{s}^{6}at)dt=e^{3\pi }.I_2....\left(iv\right)$
From Eqs. (i), (ii), (iii) and (iv), we get
$I_1=I_2+e^{\pi }.I_2+e^{2\pi }.I_2+e^{3\pi }.I_2=(1+e^{\pi }+e^{2\pi }+e^{3\pi })I_2\therefore L=\frac{{\int }_0^{4\pi }{e}^t(si{n}^{6}at+co{s}^{4}at)dt}{{\int }_0^{\pi }{e}^t(si{n}^{6}at+co{s}^{4}at)dt}=(1+e^{\pi }+e^{2\pi }+e^{3\pi })=\frac{(e^{4\pi }-1)}{e^{\pi }-1}foraϵR$