Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate \(\int_{0}^{2 \pi} \frac{x|\sin x|}{1+\cos ^{2} x} d x\). Hint: Make the substitution \(u=x-\pi\) in the definite integral and then use symmetry properties.

Step-by-Step Solution

Answer

The integral equals 0 due to the integrand's odd symmetry.

Step 1: Apply the substitution

Make the substitution \( u = x - \pi \), then \( du = dx \). Changing the limits of integration accordingly, when \( x = 0 \), \( u = -\pi \), and when \( x = 2\pi \), \( u = \pi \). Substitute in the integral to get: \[ \int_{0}^{2 \pi} \frac{x|\sin x|}{1+\cos ^{2} x} \, dx = \int_{-\pi}^{\pi} \frac{(u+\pi) |\sin(u+\pi)|}{1+\cos^2(u+\pi)} \, du\]

Step 2: Simplify using periodicity

Use the periodic properties of sine and cosine: \( \sin(u+\pi) = -\sin u \) and \( \cos(u+\pi) = -\cos u \). This gives: \[ \int_{-\pi}^{\pi} \frac{(u + \pi) |\sin u|}{1 + \cos^2 u} \, du \]

Step 3: Analyze symmetry properties

Notice that the integrand \( \frac{(u + \pi) |\sin u|}{1 + \cos^2 u} \) is an odd function about the origin since:- \((u + \pi) |\sin u|\) is odd in terms of \(u\) (as reversing \(u\) changes the sign),- \(1 + \cos^2 u\) is even.Therefore, the integrand is odd overall, and its integral from \(-\pi\) to \(\pi\) is zero by symmetry.

Key Concepts

Integration by Substitution

Integration by substitution is a method used to simplify the process of solving integrals. This technique is akin to undoing the chain rule from differentiation. By changing variables, you can transform a complex integral into a more manageable one.

In this exercise, the substitution chosen is \( u = x - \pi \). This choice helps because the integrand becomes an expression involving \( u \), which can sometimes simplify our integral significantly. After substituting, you also need to adjust the limits of integration.

• When \( x = 0 \), you find the new lower limit: \( u = 0 - \pi = -\pi \).
• When \( x = 2\pi \), the new upper limit: \( u = 2\pi - \pi = \pi \).

By performing these substitutions, you control how much the variable changes, and consequently, makes your integral easier to evaluate.

Symmetry Properties

Symmetry can be a powerful tool in calculus, particularly when evaluating definite integrals. A function is said to have a symmetrical property if it behaves predictably when its variables are inverted or offset.

In the problem provided, the function within the integral stabilizes due to its symmetry about the y-axis. By substituting \( x \) with \( u + \pi \), an odd function emerges. Odd functions have the property that \( f(-u) = -f(u) \).

• \((u + \pi) |\sin u|\) becomes odd because reversing \(u\) changes its sign.
• The denominator \(1 + \cos^2 u\) remains even, as it is symmetric about the origin.

The product of an odd function and an even function remains odd. For symmetrical intervals like \([\-\pi, \pi]\), the integral of an odd function is zero. This simplifies and saves computation while confirming the result as zero without further computation.

Trigonometric Functions

Trigonometric functions are foundational to understanding this integral. Functions like sine and cosine have specific properties, especially their periodicity and symmetry, which are useful when solving integrals.

For example, we utilize the identities:

• \(\sin(u + \pi) = -\sin u \)
• \(\cos(u + \pi) = -\cos u \)

These identities help transform the integrand after substitution by revealing properties such as oddness or evenness in terms of symmetry. This allows us to make the integral simpler and leverage symmetry properties efficiently.

Therefore, understanding how to manipulate and simplify using trigonometric identities is key. Recognizing these inherent properties accelerates problem-solving and aids in effortlessly computing or finding that an integral evaluates to zero, as in this example.