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If \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] then \[\left( {n \in N} \right)\]
A. \[n{I_1} = 5{I_2}\]
B. \[{I_1} + {I_2} = n + 5\]
C. \[\dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
D.None of these

Last updated date: 19th Jun 2024
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Hint: Here, we are required to find the relation between the given two expressions. We will use the range of \[\cos x\] and change the limit of the integral accordingly. We will find the value of the first integral and then substitute its value in the second integral. We will solve it further to get the required relation between the two expressions.

Complete step-by-step answer:
According to the question, we are given that \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]
Now, as we know that \[\cos x \in \left[ {0,\pi } \right]\].
Therefore, period of \[\left| {\cos x} \right| = \pi \]
Hence, for \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] , we can write this as:
 \[ \Rightarrow {I_1} = n\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]………………………….. \[\left( 1 \right)\]
Similarly, for \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] ,
 \[ \Rightarrow {I_2} = 5\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]
Dividing both sides by 5, we get,
 \[ \Rightarrow \dfrac{{{I_2}}}{5} = \int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] ……………………………. \[\left( 2 \right)\]
Now, substituting the value of \[\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] from equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\] , we get,
 \[{I_1} = n\dfrac{{{I_2}}}{5}\]
Dividing both sides by \[n\], we get
 \[ \Rightarrow \dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Therefore, if \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[\left( {n \in N} \right)\] , then \[\dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Hence, option C is the correct answer.

Note: In this question, we are given two definite integrals having the lower and upper limits respectively. As we know that the range of \[\cos x\] is \[\left[ {0,\pi } \right]\] this means that the minimum value which the trigonometric function \[\cos x\] can take is 0 and the maximum value taken by it can be \[\pi \]. We have shown the closed brackets because \[\cos x\] includes both the values 0 and \[\pi \]. Whereas, in the case of \[\sec x\] , its range lies between \[\left( {0,\pi } \right)\]. Hence, these open brackets show that it can never take the values 0 and \[\pi \] and hence, it will always lie between these values. Therefore, for solving this question, it is really important to know the ranges of various trigonometric functions.