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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 8 - Integration TechniquesProblem 8.3.75c
Chapter 8, Problem 8.3.75c

75. Exploring powers of sine and cosine
c. Prove that ∫₀ᵖⁱ sin²(nx) dx has the same value for all positive integers n.

Verified step by step guidance
1
Step 1: Begin by recalling the trigonometric identity for sine squared: sin²(x) = (1 - cos(2x))/2. This identity will simplify the integral.
Step 2: Substitute sin²(nx) in the integral with the identity: ∫₀ᵖⁱ sin²(nx) dx = ∫₀ᵖⁱ (1 - cos(2nx))/2 dx. Split the integral into two parts: ∫₀ᵖⁱ (1/2) dx - ∫₀ᵖⁱ (cos(2nx)/2) dx.
Step 3: Evaluate the first integral ∫₀ᵖⁱ (1/2) dx. Since this is a constant, the result is straightforward: (1/2) * x evaluated from 0 to π.
Step 4: For the second integral ∫₀ᵖⁱ (cos(2nx)/2) dx, recall that the integral of cosine over a full period (or multiples of π) is zero. Specifically, ∫₀ᵖⁱ cos(2nx) dx = 0 for any positive integer n.
Step 5: Combine the results from Step 3 and Step 4. The first integral contributes a constant value, while the second integral contributes zero. Therefore, the value of ∫₀ᵖⁱ sin²(nx) dx is independent of n and remains the same for all positive integers.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integrals

A definite integral calculates the accumulation of a function's values over a specific interval, providing the net area under the curve. In this case, the integral ∫₀ᵖⁱ sin²(nx) dx evaluates the area under the sine squared function from 0 to π, which is crucial for proving the equality for different values of n.
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Definition of the Definite Integral

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold true for all values of the variables. The identity sin²(x) = (1 - cos(2x))/2 is particularly useful in this context, as it simplifies the integral of sin²(nx) and allows for easier evaluation across different integer values of n.
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Periodicity of Sine Function

The sine function is periodic, meaning it repeats its values in regular intervals. For sin²(nx), the period is π/n, which implies that as n changes, the function still retains a consistent average value over the interval [0, π]. This periodicity is key to demonstrating that the integral yields the same result for all positive integers n.
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Related Practice
Textbook Question

Gaussians An important function in statistics is the Gaussian (or normal distribution, or bell-shaped curve), f(x) = e^(-ax²).

c. Complete the square to evaluate ∫ from -∞ to ∞ of e^(-(ax² + bx + c)) dx, where a > 0, b, and c are real numbers.

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Textbook Question

43. A hot-air balloon is launched from an elevation of 5400 ft above sea level. As it rises, the vertical velocity is computed using a device (called a variometer) that measures the change in atmospheric pressure. The vertical velocities at selected times are shown in the table (with units of ft/min).

c. A polynomial that fits the data reasonably well is:

g(t) = 3.49t³ - 43.21t² + 142.43t - 1.75

Estimate the elevation of the balloon after five minutes using this polynomial.

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Textbook Question

Prove the following orthogonality relations (which are used to generate Fourier series). Assume m and n are integers with m ≠ n.

c.

π

∫ sin(mx) cos(nx) dx = 0, when |m + n| is even

0

Textbook Question

3. What term(s) should appear in the partial fraction decomposition of a proper rational function with each of the following?

c. A factor of (x² + 2x + 6) in the denominator

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Textbook Question

94. [Use of Tech] Skydiving A skydiver has a downward velocity given by v(t) = V_T [(1 - e^(-2gt/V_T))/(1 + e^(-2gt/V_T))],

where t = 0 is the instant the skydiver starts falling, g = 9.8 m/s² is the acceleration due to gravity, and V_T is the terminal velocity of the skydiver.

c. Verify by integration that the position function is given by

s(t) = V_T t + (V_T²/g) ln[(1 + e^(-2gt/V_T))/2],

where s'(t) = v(t) and s(0) = 0.

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Textbook Question

91. [Use of Tech] Regions bounded by exponentials Let a > 0 and let R be the region bounded by the graph of y = e^(-a·x) and the x-axis

on the interval [b, ∞).

c. Find the minimum value b* such that when b > b*, there exists some a > 0 where A(a,b) = 2.