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Problem 96

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

Verified
Answer
The correct choice is (B) \(I_3 > I_1 > I_2\).
1Step 1: Evaluate Integral I3
To evaluate \(I_3 = \int_{0}^{\pi / 2} \cos x \, dx\), we can use the standard integration rule for \(\cos x\). Integrating \(\cos x\), we get \(\sin x\). Thus, \(I_3 = \left[ \sin x \right]_{0}^{\pi/2} = \sin \left( \frac{\pi}{2} \right) - \sin(0) = 1 - 0 = 1\).
2Step 2: Analyze Integral I1
For \(I_1 = \int_{0}^{\pi / 2} \cos(\sin x) \, dx\), observe that \(\cos(\sin x)\) is a positive function on \([0, \pi/2]\) and \(\sin x\) ranges from 0 to 1 in this interval. Therefore, \(\cos(\sin x)\) ranges from \(\cos(1)\) to \(\cos(0) = 1\). Thus, \(I_1\) should be less than the length of the interval times the maximum value of \(\cos(\sin x)\), i.e., \(I_1 < \frac{\pi}{2} \times 1 < 1\).
3Step 3: Analyze Integral I2
For \(I_2 = \int_{0}^{\pi / 2} \sin(\cos x) \, dx\), note that \(\cos x\) decreases from 1 to 0 as \(x\) goes from 0 to \(\pi/2\). Therefore, \(\sin(\cos x)\) ranges from \(\sin(1)\) to \(\sin(0) = 0\). Since \(\sin(1) < 1\), \(I_2\) is also less than 1. Additionally, as \(\sin(\cos x)\) is generally smaller than \(\cos(\sin x)\), \(I_2 < I_1\).
4Step 4: Compare the Values
With the evaluations and inequalities determined: \(I_3 = 1\), \(I_1 < 1\), and \(I_2 < I_1\), we conclude that \(I_3 > I_1 > I_2\).

Key Concepts

Integral InequalityTrigonometric FunctionsCalculus Problem Solving
Integral Inequality
Understanding integral inequalities can help solve challenging calculus problems. In this exercise, we have three different integrals, each representing a definite area under a curve on the interval from 0 to \(\frac{\pi}{2}\). By comparing these integrals, we utilize the concept of inequalities to determine their relative sizes.

Each integral gives us a number, which represents the area enclosed by their respective curves and the x-axis over the given interval. By evaluating each integrand's behavior, we gain insights about the size of each integral without fully calculating all their exact values.
  • I_3=∫cos⁡x space (콜로>ddxmo=1l>+l>
  • I_1<∫c(ml)))>⁡을math>xddx이math='1'>남다.
  • I_2<I_ ＜1 >1=xOML" auto>⁡이 떠모>morel>These inequalities help us conclude which integrals have the largest and smallest areas. This insight into integral ranking is practical for calculus problem-solving.
    Trigonometric Functions
    Trigonometric functions play a crucial role in calculus and in solving the given problem. Here, the functions \(\cos(\sin x)\) and \(\sin(\cos x)\) are evaluated over the interval \([0, \frac{\pi}{2}]\). This ranges in trigonometric functions aren't standard and need a careful exploration.

    • For \(\cos(\sin x)\), as \(x\) advances from \(0\) to \(\frac{\pi}{2}\), \(\sin x\) varies from \(0\) to \(1\). Therefore, \(\cos(\sin x)\) ranges from \(\cos(1)\) to \(\cos(0) = 1\).
    • In contrast, for \(\sin(\cos x)\), \(\cos x\) decreases from \(1\) to \(0\) as \(x\) grows. Thus, \(\sin(\cos x)\) ranges from \(\sin(1)\) to \(\sin(0) = 0\).

    Both functions require a mental map of how trigonometric values shift over this specific interval. This understanding is fundamental when estimating the integrals and comparing their areas.
    Calculus Problem Solving
    Calculus problem-solving often involves evaluating integrals and applying concepts such as inequalities and trigonometry, as demonstrated in this exercise. We combined these techniques to deduce the relationship between \(I_1\), \(I_2\), and \(I_3\).

    To begin, calculating \(I_3\) accurately, we determine it as \(1\) through direct integration, applying:
    • Integration rule: \(\int \cos(x) \, dx = \sin(x)\)
    • Definite integral evaluation: \(\left[ \sin(x) \right]_{0}^{\frac{\pi}{2}} = 1 - 0 = 1\)
    Next, for \(I_1\) and \(I_2\), recognizing and comparing the behavior and maximum values of the integrands over \([0, \frac{\pi}{2}]\) allows us to establish inequalities:
    • \(I_1 < 1\) because \(\cos(\sin x)\) caps at \(1\)
    • \(I_2 < I_1\), since \(\sin(\cos x)\) achieves lower maxima

    Consequently, these observations allowed us to deduce that \(I_3 > I_1 > I_2\). Carefully assessing these functions and inequalities leads us to the correct relationship and solution.
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