Problem 41 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(I_{1}=\int_{0}^{3 \pi} f\left(\cos ^{2} x\right) d x\) and \(I_{2}=\int_{0}^{\pi} f\left(\cos ^{2} x\right) d x\), then (A) \(I_{1}=5 I_{2}\) (B) \(I_{1}=I_{2}\) (C) \(I_{1}=3 I_{2}\) (D) None of these

Step-by-Step Solution

Answer

\(I_{1}=3 I_{2}\), so the answer is (C).

Step 1: Understanding the Given Integrals

We have two integrals, \( I_1 = \int_0^{3\pi} f(\cos^2 x) dx \) and \( I_2 = \int_0^{\pi} f(\cos^2 x) dx \). Our goal is to determine the relationship between \( I_1 \) and \( I_2 \).

Step 2: Analyzing the Function

Observe that \( f(\cos^2 x) \) is periodic with respect to its argument. The function \( \cos^2 x \) has a period of \( \pi \) since it repeats every \( \pi \). This is due to the periodic nature of the cosine function.

Step 3: Decomposing the Integral I1

Since \( \cos^2 x \) has a period of \( \pi \), we can break the integral \( I_1 \) into three intervals: \([0, \pi], [\pi, 2\pi], [2\pi, 3\pi]\). Thus, \( I_1 = \int_0^{\pi} f(\cos^2 x) dx + \int_{\pi}^{2\pi} f(\cos^2 x) dx + \int_{2\pi}^{3\pi} f(\cos^2 x) dx \).

Step 4: Evaluating Each Part of I1

Each interval integral, \( \int_0^{\pi} f(\cos^2 x) dx \), \( \int_{\pi}^{2\pi} f(\cos^2 x) dx \), and \( \int_{2\pi}^{3\pi} f(\cos^2 x) dx \), evaluates to \( I_2 \) because the function \( f(\cos^2 x) \) has the same effect over these intervals due to periodicity. Thus, \( I_1 = I_2 + I_2 + I_2 = 3I_2 \).

Step 5: Conclusion

Based on our analysis, the relationship between \( I_1 \) and \( I_2 \) is \( I_1 = 3I_2 \). The correct answer is option C.

Key Concepts

Periodic Functions

The concept of periodic functions is fundamental when dealing with calculus and integrals. A periodic function is one that repeats its values in regular intervals or periods. In mathematical terms, a function \( f(x) \) is periodic with period \( T \) if \( f(x + T) = f(x) \) for all \( x \). Understanding this repetition is crucial for simplifying complex integrals over a wider range.
In the given problem, the function involved is \( \cos^2 x \), which is a periodic function with a period of \( \pi \). This means every \( \pi \) interval, the function \( \cos^2 x \) starts to repeat its values. Hence, any function of \( \cos^2 x \), such as \( f(\cos^2 x) \), will also exhibit periodicity over the same interval.

• Periodic functions simplify calculations.
• They allow breakdown of integrals into segments of the period.
• Repetition within each segment results in equivalent integral values.

This periodic nature permits us to break down the integral over multiple periods, as demonstrated in the solution.

Cosine Function

The cosine function is a fundamental trigonometric function that oscillates between -1 and 1. It is an even function, meaning \( \cos(-x) = \cos(x) \), and has a period of \( 2\pi \). This property is the foundation for understanding more complex periodic functions, like \( \cos^2 x \).

When the cosine function is squared, \( \cos^2 x \), the period becomes \( \pi \). This occurs because squaring removes the negative half of the cycle, effectively mirroring each cycle into positive values only. Thus, the repetition happens every \( \pi \) rather than \( 2\pi \).

This property of the cosine function helps determine the periodicity of \( \cos^2 x \), crucial for evaluating integrals by allowing breakdown into equal, manageable segments over integer multiples of \( \pi \).

• \( \cos^2 x \) has a smaller period due to squaring.
• Useful in splitting integrals across phases.
• Simplifies evaluation of definite integrals.

Integral Properties

Understanding the properties of integrals is key to solving definite integrals, especially when working with periodic functions. One such property is that if a function is periodic with period \( T \), the integral over a full cycle \( [0, T] \) will capture the entire repeating pattern. Therefore, integrals over intervals comprising whole numbers of periods can be expressed in terms of one such integral.

For instance, integral property helps us write the integral over three periods \( [0, 3\pi] \) in terms of an integral over one period \( [0, \pi] \). This decomposition is possible because the function repeats every \( \pi \), giving us:

• \( I_1 = 3 \times I_2 \), where each entire period contributes equally.

Recognizing and utilizing these properties ensure effective breakdown of complex integration problems into simpler, solvable parts, as seen in this solution.