Problem 96 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(I_{1}=\int_{0}^{\pi / 2} \cos (\sin x) d x ; I_{2}=\int_{0}^{\pi / 2} \sin (\cos x) d x\) and \(I_{3}=\int_{0}^{\pi / 2} \cos x d x\), then (A) \(I_{1}>I_{3}>I_{2}\) (B) \(I_{3}>I_{1}>I_{2}\) (C) \(I_{1}>I_{2}>I_{3}\) (D) \(I_{3}>I_{2}>I_{1}\)

Step-by-Step Solution

Answer

Option (B), \(I_3 > I_1 > I_2\).

Step 1: Evaluate the Integral I3

The integral \( I_3 = \int_{0}^{\pi / 2} \cos x \, dx \) can be evaluated directly. It is a standard integral where the antiderivative of \( \cos x \) is \( \sin x \). Thus, \( I_3 = [ \sin x ]_{0}^{\pi / 2} = \sin(\pi / 2) - \sin(0) = 1 - 0 = 1 \).

Step 2: Consider the Integrals I1 and I2

Both integrals \( I_1 \) and \( I_2 \) are special functions involving trigonometric compositions and do not have elementary antiderivatives. To compare their values with \( I_3 \,=\, 1 \), qualitative analysis and inequalities will be used.

Step 3: Analyze Inequalities for I1

For \( I_1 = \int_{0}^{\pi / 2} \cos(\sin x) \, dx \), notice that \( 0 \leq \sin x \leq 1 \) for \( x \) in the range \([0, \pi/2] \). Thus, \( \cos(\sin x) \geq \cos(1) \) and the integral is approximately greater than \( \cos(1) \times \frac{\pi}{2} \), which makes it less than \( I_3 \).

Step 4: Analyze Inequalities for I2

For \( I_2 = \int_{0}^{\pi / 2} \sin(\cos x) \, dx \), we know \( \cos x \) decreases from 1 to 0 over the range \([0, \pi/2] \) meaning the maximum value of \( \sin(\cos x) \) is 1. The average effect leads to \( I_2 \) typically being smaller than \( I_3 \) and intuitively smaller than \( I_1 \) because of dominance by small values of \( \sin \).

Step 5: Compare and Conclude

From the analysis: \( I_3 = 1 \) is greater than \( I_1 \), which is greater than \( I_2 \), thus the correct order in inequalities is \( I_3 > I_1 > I_2 \).

Key Concepts

Trigonometric Functions

Trigonometric functions play a key role in mathematics, especially when analyzing integrals involving angles and periodic relationships. In calculus, these functions often appear within integrals related to oscillating patterns and cycles, such as wave forms and circular motions.
When dealing with trigonometric integrals, like in the exercise provided, we encounter functions such as \( \sin(x)\) and \( \cos(x)\). These functions:

• Represent angles and rotations.
• Have well-defined derivatives and integrals, making them fundamental in calculus problems.
• Allow us to understand periodic and wave-like phenomena, which are essential in fields ranging from physics to engineering.

The use of compositions, seen in functions like \( \cos(\sin \ x)\) and \( \sin(\cos \ x)\), introduces complexity as they combine two trigonometric functions, leading to no straightforward antiderivative. These combinations require careful analysis using other methods, such as numerical integration, comparison, or graphical analysis.

Inequality Analysis

Inequality analysis is a powerful tool in evaluating and comparing expressions when direct calculation isn't feasible. In the context of integrals, inequalities help in estimating bounds and understanding the relative size of different definite integrals.
For our problem, inequality analysis was crucial in comparing \( I_1, \ I_2\), and \( I_3\). Here's how we used it:

• For \( I_1 = \int_{0}^{\pi / 2} \cos(\sin \ x) \ dx \), recognizing that \( 0 \leq \sin x \leq \1 \) means \( \cos(\sin x) \geq \cos(1)\), giving an approximate lower bound for the integral.
• For \( I_2 = \int_{0}^{\pi / 2} \sin(\cos \ x) \ dx \), understanding \( \cos x \) varies from 1 to 0 implies the maximum value \( \sin(\cos x) \) can take is 1. Due to the properties of sine and cosine, \( I_2 \) tends to be smaller compared to others.

Inequalities provide insight without needing exact solutions, which is particularly handy when dealing with complex integrals of non-standard functions.

Antiderivatives

Antiderivatives, known as indefinite integrals, reverse the operation of differentiation. Finding an antiderivative is crucial in calculating definite integrals. Essentially, it involves identifying a function whose derivative yields the given function.
For common functions, like \( \cos x \), the antiderivative is easy to find, as demonstrated when calculating \( I_3 \) where \( \sin x \) is the antiderivative of \( \cos x \). However, more complex functions like \( \cos(\sin x) \) or \( \sin(\cos x) \) lack elementary antiderivatives. This means:

• They can't be integrated using the basic rules of integration.
• We might resort to numerical methods or approximation techniques.
• We apply strategies like substitution or series expansion in some cases.

Understanding antiderivatives gives us a solid foundation for efficiently solving definite integrals, crucial for quantitative problems.

Integral Comparison

Integral comparison is an effective method to deduce the relative sizes of integrals, especially when direct evaluation is complex or impossible. By establishing relationships between integrals, we make informed decisions about their outcomes.
In the exercise provided, we compared three integrals:

• \( I_3 = 1 \), due to its straightforward trigonometric nature, which serves as a reference point.
• We determined \( I_1 \), involving \( \cos(\sin x) \), where the function's value is above that of the function in \( I_2 \), proving \( I_1 \) is larger than \( I_2 \).
• Observing that \( I_3 \), being a simple standard function, proved to be greater than both \( I_1 \) and \( I_2 \) due to their more constrained functional forms.

Integral comparison reveals hierarchy between integrals, offering clarity when precise solutions are unavailable, and ensuring confidence in rankings based on intuitive and rigorous reasoning.