The options with the values of $\alpha $ and $L$ that satisfies the following equation is/are
$\dfrac{\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}{\int\limits_{0}^{\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}=L,$ is/are
A. $\alpha =2,L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
B. $\alpha =2,L=\dfrac{{{e}^{4\pi }}+1}{{{e}^{\pi }}+1}$
C. $\alpha =4,L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
D. $\alpha =4,L=\dfrac{{{e}^{4\pi }}+1}{{{e}^{\pi }}+1}$
Answer
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Hint: We solve this question by assuming the value of $\alpha $ as 2 and 4 and solving for $L.$ We use the basic integration formula and concepts to simplify the expression. We split the numerator to intervals of $\pi $ and simplify this function in such a way that the numerator can be simplified to obtain the solution.
Complete step by step solution:
In order to solve this question, let us consider the numerator of the given function $L=\dfrac{\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}{\int\limits_{0}^{\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}.\ldots \left( 1 \right)$
Let us consider this numerator to be I.
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}$
Let us consider $\alpha =2$ in the first case,
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}2t+{{\cos }^{4}}2t \right)dt}$
Let us split this into 4 intervals each of interval $\pi .$ Also consider $f\left( t \right)={{e}^{t}}\left( {{\sin }^{6}}2t+{{\cos }^{4}}2t \right).$
$\Rightarrow I=\int\limits_{0}^{\pi }{f\left( t \right)dt}+\int\limits_{\pi }^{2\pi }{f\left( t \right)dt}+\int\limits_{2\pi }^{3\pi }{f\left( t \right)dt}+\int\limits_{3\pi }^{4\pi }{f\left( t \right)dt}$
Let us assume each of this term as ${{I}_{1}},{{I}_{2}},{{I}_{3}},{{I}_{4}}$ respectively.
$\Rightarrow I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}$
For ${{I}_{1}},$ let $t=x,$ and $dt=dx.$ Hence, ${{I}_{1}}$ is given by $\int\limits_{0}^{\pi }{f\left( x \right)dx}.$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Consider ${{I}_{2}}$ and substitute $t=\pi +x.$ Differentiating this,
$\Rightarrow dt=dx$
Because of this, the limits change from 0 to $\pi .$ Therefore, ${{I}_{2}}$ can be written as,
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{f\left( \pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{{{e}^{x+\pi }}\left( {{\sin }^{6}}2\left( x+\pi \right)+{{\cos }^{4}}2\left( x+\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+\pi }}$ can be represented as ${{e}^{x}}.{{e}^{\pi }}.$ This ${{e}^{\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{2}}={{e}^{\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Similarly, for the term ${{I}_{3}},$ let us consider $t=2\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{3}}$ can be written as,
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{f\left( 2\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{{{e}^{x+2\pi }}\left( {{\sin }^{6}}2\left( x+2\pi \right)+{{\cos }^{4}}2\left( x+2\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+2\pi }}$ can be represented as ${{e}^{x}}.{{e}^{2\pi }}.$ This ${{e}^{2\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{3}}={{e}^{2\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Similarly, for the term ${{I}_{4}},$ let us consider $t=3\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi.$ Therefore, ${{I}_{4}}$ can be written as,
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{f\left( 3\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{{{e}^{x+3\pi }}\left( {{\sin }^{6}}2\left( x+3\pi \right)+{{\cos }^{4}}2\left( x+3\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+3\pi }}$ can be represented as ${{e}^{x}}.{{e}^{3\pi }}.$ This ${{e}^{3\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{4}}={{e}^{3\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Substituting this in the equation 1 for the numerator and denominator,
$\Rightarrow L=\dfrac{I}{{{I}_{1}}}=\dfrac{{{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}}{{{I}_{1}}}$
Substituting for these values and taking the term common out from the numerator,
$\Rightarrow L=\dfrac{\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}}{\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}}$
Cancelling the common terms in numerator and denominator,
$\Rightarrow L=\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)$
We know that the geometric progression can be written in a general form as $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},$ where a is the first term, n is the number of terms, r is the common ratio. We represent L in this form,
$\Rightarrow L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
Hence, option A is correct.
Similarly, we solve for $\alpha =4.$ Let us consider the numerator term as $I.$
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}4t+{{\cos }^{4}}4t \right)dt}$
Let us split this into 4 intervals each of interval $\pi .$ Also consider $f\left( t \right)={{e}^{t}}\left( {{\sin }^{6}}4t+{{\cos }^{4}}4t \right).$
$\Rightarrow I=\int\limits_{0}^{\pi }{f\left( t \right)dt}+\int\limits_{\pi }^{2\pi }{f\left( t \right)dt}+\int\limits_{2\pi }^{3\pi }{f\left( t \right)dt}+\int\limits_{3\pi }^{4\pi }{f\left( t \right)dt}$
Let us assume each of this term as ${{I}_{1}},{{I}_{2}},{{I}_{3}},{{I}_{4}}$ respectively.
$\Rightarrow I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}$
For ${{I}_{1}},$ let $t=x,$ and $dt=dx.$ Hence, ${{I}_{1}}$ is given by $\int\limits_{0}^{\pi }{f\left( x \right)dx}.$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Consider ${{I}_{2}}$ and substitute $t=\pi +x.$ Differentiating this,
$\Rightarrow dt=dx$
Because of this, the limits change from 0 to $\pi .$ Therefore, ${{I}_{2}}$ can be written as,
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{f\left( \pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{{{e}^{x+\pi }}\left( {{\sin }^{6}}4\left( x+\pi \right)+{{\cos }^{4}}4\left( x+\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+\pi }}$ can be represented as ${{e}^{x}}.{{e}^{\pi }}.$ This ${{e}^{\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{2}}={{e}^{\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Similarly, for the term ${{I}_{3}},$ let us consider $t=2\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{3}}$ can be written as,
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{f\left( 2\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{{{e}^{x+2\pi }}\left( {{\sin }^{6}}4\left( x+2\pi \right)+{{\cos }^{4}}4\left( x+2\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+2\pi }}$ can be represented as ${{e}^{x}}.{{e}^{2\pi }}.$ This ${{e}^{2\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{3}}={{e}^{2\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Similarly, for the term ${{I}_{4}},$ let us consider $t=3\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{4}}$ can be written as,
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{f\left( 3\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{{{e}^{x+3\pi }}\left( {{\sin }^{6}}4\left( x+3\pi \right)+{{\cos }^{4}}4\left( x+3\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+3\pi }}$ can be represented as ${{e}^{x}}.{{e}^{3\pi }}.$ This ${{e}^{3\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{4}}={{e}^{3\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Substituting this in the equation 1 for the numerator and denominator,
$\Rightarrow L=\dfrac{I}{{{I}_{1}}}=\dfrac{{{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}}{{{I}_{1}}}$
Substituting for these values and taking the term common out from the numerator,
$\Rightarrow L=\dfrac{\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}}{\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}}$
Cancelling the common terms in numerator and denominator,
$\Rightarrow L=\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)$
We know that the geometric progression can be written in a general form as $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},$ where a is the first term, n is the number of terms, r is the common ratio. We represent L in this form,
$\Rightarrow L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
Hence, option C is the other correct option. Hence, the two correct options are A and C.
Note:
We need to know the basics of integration limits and their conversions. We need to be careful while changing the variable, we need to change the limits too. Students must be careful while considering the geometric progression too.
Complete step by step solution:
In order to solve this question, let us consider the numerator of the given function $L=\dfrac{\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}{\int\limits_{0}^{\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}}.\ldots \left( 1 \right)$
Let us consider this numerator to be I.
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}\alpha t+{{\cos }^{4}}\alpha t \right)dt}$
Let us consider $\alpha =2$ in the first case,
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}2t+{{\cos }^{4}}2t \right)dt}$
Let us split this into 4 intervals each of interval $\pi .$ Also consider $f\left( t \right)={{e}^{t}}\left( {{\sin }^{6}}2t+{{\cos }^{4}}2t \right).$
$\Rightarrow I=\int\limits_{0}^{\pi }{f\left( t \right)dt}+\int\limits_{\pi }^{2\pi }{f\left( t \right)dt}+\int\limits_{2\pi }^{3\pi }{f\left( t \right)dt}+\int\limits_{3\pi }^{4\pi }{f\left( t \right)dt}$
Let us assume each of this term as ${{I}_{1}},{{I}_{2}},{{I}_{3}},{{I}_{4}}$ respectively.
$\Rightarrow I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}$
For ${{I}_{1}},$ let $t=x,$ and $dt=dx.$ Hence, ${{I}_{1}}$ is given by $\int\limits_{0}^{\pi }{f\left( x \right)dx}.$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Consider ${{I}_{2}}$ and substitute $t=\pi +x.$ Differentiating this,
$\Rightarrow dt=dx$
Because of this, the limits change from 0 to $\pi .$ Therefore, ${{I}_{2}}$ can be written as,
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{f\left( \pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{{{e}^{x+\pi }}\left( {{\sin }^{6}}2\left( x+\pi \right)+{{\cos }^{4}}2\left( x+\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+\pi }}$ can be represented as ${{e}^{x}}.{{e}^{\pi }}.$ This ${{e}^{\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{2}}={{e}^{\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Similarly, for the term ${{I}_{3}},$ let us consider $t=2\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{3}}$ can be written as,
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{f\left( 2\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{{{e}^{x+2\pi }}\left( {{\sin }^{6}}2\left( x+2\pi \right)+{{\cos }^{4}}2\left( x+2\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+2\pi }}$ can be represented as ${{e}^{x}}.{{e}^{2\pi }}.$ This ${{e}^{2\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{3}}={{e}^{2\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Similarly, for the term ${{I}_{4}},$ let us consider $t=3\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi.$ Therefore, ${{I}_{4}}$ can be written as,
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{f\left( 3\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{{{e}^{x+3\pi }}\left( {{\sin }^{6}}2\left( x+3\pi \right)+{{\cos }^{4}}2\left( x+3\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 2x. Also, the ${{e}^{x+3\pi }}$ can be represented as ${{e}^{x}}.{{e}^{3\pi }}.$ This ${{e}^{3\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{4}}={{e}^{3\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}$
Substituting this in the equation 1 for the numerator and denominator,
$\Rightarrow L=\dfrac{I}{{{I}_{1}}}=\dfrac{{{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}}{{{I}_{1}}}$
Substituting for these values and taking the term common out from the numerator,
$\Rightarrow L=\dfrac{\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}}{\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}2x+{{\cos }^{4}}2x \right)dx}}$
Cancelling the common terms in numerator and denominator,
$\Rightarrow L=\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)$
We know that the geometric progression can be written in a general form as $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},$ where a is the first term, n is the number of terms, r is the common ratio. We represent L in this form,
$\Rightarrow L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
Hence, option A is correct.
Similarly, we solve for $\alpha =4.$ Let us consider the numerator term as $I.$
$\Rightarrow I=\int\limits_{0}^{4\pi }{{{e}^{t}}\left( {{\sin }^{6}}4t+{{\cos }^{4}}4t \right)dt}$
Let us split this into 4 intervals each of interval $\pi .$ Also consider $f\left( t \right)={{e}^{t}}\left( {{\sin }^{6}}4t+{{\cos }^{4}}4t \right).$
$\Rightarrow I=\int\limits_{0}^{\pi }{f\left( t \right)dt}+\int\limits_{\pi }^{2\pi }{f\left( t \right)dt}+\int\limits_{2\pi }^{3\pi }{f\left( t \right)dt}+\int\limits_{3\pi }^{4\pi }{f\left( t \right)dt}$
Let us assume each of this term as ${{I}_{1}},{{I}_{2}},{{I}_{3}},{{I}_{4}}$ respectively.
$\Rightarrow I={{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}$
For ${{I}_{1}},$ let $t=x,$ and $dt=dx.$ Hence, ${{I}_{1}}$ is given by $\int\limits_{0}^{\pi }{f\left( x \right)dx}.$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{1}}=\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Consider ${{I}_{2}}$ and substitute $t=\pi +x.$ Differentiating this,
$\Rightarrow dt=dx$
Because of this, the limits change from 0 to $\pi .$ Therefore, ${{I}_{2}}$ can be written as,
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{f\left( \pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{2}}=\int\limits_{0}^{\pi }{{{e}^{x+\pi }}\left( {{\sin }^{6}}4\left( x+\pi \right)+{{\cos }^{4}}4\left( x+\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+\pi }}$ can be represented as ${{e}^{x}}.{{e}^{\pi }}.$ This ${{e}^{\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{2}}={{e}^{\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Similarly, for the term ${{I}_{3}},$ let us consider $t=2\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{3}}$ can be written as,
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{f\left( 2\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{3}}=\int\limits_{0}^{\pi }{{{e}^{x+2\pi }}\left( {{\sin }^{6}}4\left( x+2\pi \right)+{{\cos }^{4}}4\left( x+2\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+2\pi }}$ can be represented as ${{e}^{x}}.{{e}^{2\pi }}.$ This ${{e}^{2\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{3}}={{e}^{2\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Similarly, for the term ${{I}_{4}},$ let us consider $t=3\pi +x.$ For this again, $dt=dx$ and the limits change from 0 to $\pi .$ Therefore, ${{I}_{4}}$ can be written as,
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{f\left( 3\pi +x \right)dx}$
Substituting for $f\left( x \right),$
$\Rightarrow {{I}_{4}}=\int\limits_{0}^{\pi }{{{e}^{x+3\pi }}\left( {{\sin }^{6}}4\left( x+3\pi \right)+{{\cos }^{4}}4\left( x+3\pi \right) \right)dx}$
Multiplying the term in brackets, cos and sin repeat after every multiple of $2\pi $ values, hence we can just write the term as 4x. Also, the ${{e}^{x+3\pi }}$ can be represented as ${{e}^{x}}.{{e}^{3\pi }}.$ This ${{e}^{3\pi }}$ can be taken outside the integral since it’s a constant.
$\Rightarrow {{I}_{4}}={{e}^{3\pi }}\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}$
Substituting this in the equation 1 for the numerator and denominator,
$\Rightarrow L=\dfrac{I}{{{I}_{1}}}=\dfrac{{{I}_{1}}+{{I}_{2}}+{{I}_{3}}+{{I}_{4}}}{{{I}_{1}}}$
Substituting for these values and taking the term common out from the numerator,
$\Rightarrow L=\dfrac{\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}}{\int\limits_{0}^{\pi }{{{e}^{x}}\left( {{\sin }^{6}}4x+{{\cos }^{4}}4x \right)dx}}$
Cancelling the common terms in numerator and denominator,
$\Rightarrow L=\left( 1+{{e}^{\pi }}+{{e}^{2\pi }}+{{e}^{3\pi }} \right)$
We know that the geometric progression can be written in a general form as $\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1},$ where a is the first term, n is the number of terms, r is the common ratio. We represent L in this form,
$\Rightarrow L=\dfrac{{{e}^{4\pi }}-1}{{{e}^{\pi }}-1}$
Hence, option C is the other correct option. Hence, the two correct options are A and C.
Note:
We need to know the basics of integration limits and their conversions. We need to be careful while changing the variable, we need to change the limits too. Students must be careful while considering the geometric progression too.
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