Problem 50 - 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

(B) \( I_3>I_1>I_2 \)

Step 1: Evaluate I_3

Start with the simplest integral, \[ I_3 = \int_{0}^{\pi / 2} \cos x \, dx \]. The antiderivative of \( \cos x \) is \( \sin x \). Applying the limits of integration, we have:\[I_3 = \left[ \sin x \right]_{0}^{\pi / 2} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1.\]

Step 2: Evaluate I_1 and I_2 using Comparison

To estimate \(I_1\) and \(I_2\), observe the behavior of the functions \(\cos(\sin x)\) and \(\sin(\cos x)\) within the interval \([0, \pi/2]\). For \(x\) in \([0, \pi/2]\), \(\sin x\) ranges from 0 to 1 and \(\cos x\) ranges from 1 to 0.\(\cos(\sin x)\) tends to be slightly less than 1 and \(\sin(\cos x)\) tends to be slightly less than \(\sin(1)\). Both functions are typically less than 1, implying:\[ I_3 > I_1 \, \text{and} \, I_3 > I_2.\] Also, because \(\cos(\sin x) > \sin(\cos x)\), it follows that \(I_1 > I_2\).

Step 3: Determine the Order

Based on the previous evaluations: - \(I_3 = 1\)- \(I_3 > I_1\)- \(I_3 > I_2\)- \(I_1 > I_2\)This implies the order\[ I_3 > I_1 > I_2. \]

Key Concepts

Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent are fundamental in mathematics, particularly in the study of triangles and modelling periodic phenomena.
In the context of definite integrals, they appear frequently due to their periodic nature and the circular motion they describe.
For example, in the given problem, we examine integrals involving functions like \( \cos(\sin x) \) and \( \sin(\cos x) \).
Trigonometric functions can transform angles into linear values, which makes them versatile in integrating over specific intervals.
They map periodic inputs (like angles) to outputs that are bound between certain values, such as -1 and 1 for sine and cosine.

• \( \cos(\sin x) \) is interesting because it combines two trigonometric functions into a composite function, first calculating the sine of \( x \), then using that result as the input for cosine.
• \( \sin(\cos x) \) does the reverse, where cosine of \( x \) becomes the input for sine, altering the output range.

Understanding these functions and their compositions is crucial, as they affect the behavior of the integrals formed with them, influencing how we approach their evaluation across defined intervals.

Integral Calculus

Integral calculus is a major branch of calculus focused on the concept of accumulation, enabling us to calculate areas, volumes, and totals.
A definite integral, such as those evaluated here, often represents the area under a curve between two specific limits.
In integral calculus, finding the anti-derivative or primitive is key to evaluating definite integrals.
This process is depicted in the example solution where \( I_3 = \int_0^{\pi/2} \cos x \, dx \) is evaluated.

• First, identify the primitive function of \( \cos x \), which is \( \sin x \).
• Then, apply the fundamental theorem of calculus to evaluate: \( \left[ \sin x \right]_{0}^{\pi/2} \).
• Simplifying the expression gives the result 1, reflecting the complete area under \( \cos x \) from 0 to \( \pi/2 \).

These methods enable us to understand and solve more complex integrals involving trigonometric functions and other composite functions within defined boundaries.

Comparison of Integrals

Integrals can be compared to understand the relative sizes of calculated areas, which helps in ranking them accordingly.
In this exercise, we compare \( I_1 = \int_0^{\pi/2} \cos(\sin x) \, dx \), \( I_2 = \int_0^{\pi/2} \sin(\cos x) \, dx \), and \( I_3 = \int_0^{\pi/2} \cos x \, dx \) to determine their relative magnitudes.

• Start by evaluating \( I_3 \), resulting in 1, as shown in the solution.
• Next, we note that both \( \cos(\sin x) \) and \( \sin(\cos x) \) are less than 1 over \( [0, \pi/2] \), indicating that their full integrals are less than 1.
• The relationship \( \cos(\sin x) > \sin(\cos x) \) suggests \( I_1 > I_2 \).
• Thus, the comparison order is established as \( I_3 > I_1 > I_2 \).

Such comparisons are valuable in calculus as they help underline differences in function behaviors and their impacts on integrated results.