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Chapter 7: Problem 35

Evaluate the integral. $$ \int_{-\pi / 2}^{\pi / 2} \frac{x}{1+\cos ^{2} x} d x $$

Short Answer

Expert verified
The integral evaluates to 0 because the integrand is odd over a symmetric interval.

Step by step solution

01

Set up the Integration Problem

We need to evaluate the definite integral \( \int_{-\pi / 2}^{\pi / 2} \frac{x}{1+\cos ^{2} x} \, dx \). The integrand is \( \frac{x}{1+\cos^2 x} \), and the limits of integration are from \(-\pi/2\) to \(\pi/2\).
02

Check for Symmetric Properties

Notice that the integrand function \( \frac{x}{1+\cos^2 x} \) is odd due to the presence of \(x\) in the numerator and even function \(\cos^2 x\) in the denominator. If \( f(-x) = -f(x) \), the integral from \(-a\) to \(a\) of an odd function is zero.
03

Verify the Function is Odd

Let's substitute \(-x\) into the integrand: \( f(-x) = \frac{-x}{1+\cos^2(-x)} = \frac{-x}{1+\cos^2 x} \). Clearly, \( f(-x) = -f(x) \), confirming that the function is odd.
04

Use Symmetry to Evaluate the Integral

Since \( \frac{x}{1+\cos^2 x} \) is odd and we are integrating over a symmetric interval \([-\pi/2, \pi/2]\), we find that \( \int_{-\pi/2}^{\pi/2} f(x) \, dx = 0 \).
05

Conclude the Solution

The integral \( \int_{-\pi/2}^{\pi/2} \frac{x}{1+\cos^2 x} \, dx \) evaluates to zero because the integrand is an odd function over a symmetric interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral is a fundamental concept in calculus. It represents the accumulation of quantities, like area under a curve, between two specified points. In notation, it looks like this: \[ \int_{a}^{b} f(x) \, dx \]
  • \( a \) and \( b \) are the limits of integration, showing the interval over which you integrate.
  • \( f(x) \) is the function you are integrating.
  • The results help find the net area between the curve and the x-axis over \( [a, b] \).
In our problem, the definite integral \( \int_{-\pi/2}^{\pi/2} \frac{x}{1+\cos^2 x} \, dx \) provides the net contribution of the function \( \frac{x}{1+\cos^2 x} \) over the interval from \(-\pi/2\) to \(\pi/2\). Calculating definite integrals helps in various fields like physics, engineering, and economics to find exact measures.
Odd Function
An odd function is one where substituting \(-x\) for \(x\) yields the negative of the original function: \[ f(-x) = -f(x) \]
  • Odd functions exhibit symmetry around the origin in a Cartesian coordinate system.
  • Graphically, if you rotate an odd function's graph 180 degrees about the origin, it looks unchanged.
  • Common examples include \( f(x) = x^3 \) and \( f(x) = \sin x \).
In the step-by-step solution, we confirmed that \( \frac{x}{1+\cos^2 x} \) is an odd function, which simplifies the integration process when dealing with symmetric intervals.
Symmetry in Integration
Symmetry plays a crucial role in simplifying calculations of integrals. When dealing with definite integrals, symmetry can make the process much faster:
  • If a function is odd and symmetric around zero, integrating over symmetrical intervals like \([-a, a]\) results in zero.
  • This is because the areas under the curve on either side of the origin cancel each other out.
  • The concept greatly aids in verifying the properties of functions, reducing computational overhead.
Applying symmetry in our integral \( \int_{-\pi/2}^{\pi/2} \frac{x}{1+\cos^2 x} \, dx \) was key. Since the function was odd, the integral over the symmetric interval \([-\pi/2, \pi/2]\) conveniently resulted in a value of zero.
Trigonometric Functions
Trigonometric functions encompass ratios of sides in a right triangle, applied in a circular context. They are foundational in describing periodic phenomena. For example:
  • \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are basic trigonometric functions.
  • \( \cos^2(x) \) refers to \((\cos(x))^2\), often used in integrals.
In our discussion, \( \cos^2 x \) appears in the denominator of the function being integrated. Understanding these functions helps to analyze periodic behavior and solve complex integrals efficiently.

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