Question

Perform the indicated integrations. \(\int_{0}^{\pi} \frac{\pi x-1}{\sqrt{x^{2}+\pi^{2}}} d x\)

Short answer

The integral evaluates to 0 due to symmetry.

Step-by-step solution

Identify the Integration Technique

The given integral is \( \int_{0}^{\pi} \frac{\pi x - 1}{\sqrt{x^{2} + \pi^{2}}} \, dx \). The integrand is a rational function divided by a square root, which suggests that substitution or parts might be involved. However, due to its complexity, it's often beneficial to look at it as a problem where symmetry or identity simplifications could simplify the evaluation process.

Check for Symmetry

The limits of integration range from 0 to \( \pi \), and the integrand involves a term \( \pi x - 1 \). This can suggest symmetry might help. Consider transforming the variable by setting \( u = \pi - x \). This simplifies and transforms our integral into another equivalent integral based on similar boundaries.

Variable Substitution

Let \( u = \pi - x \). Then \( du = -dx \). The new limits when \( x = 0 \), \( u = \pi \); when \( x = \pi \), \( u = 0 \). Thus, the integral becomes:\[\int_{\pi}^{0} \frac{(\pi (\pi-u) - 1)}{\sqrt{(\pi-u)^{2} + \pi^{2}}} \,(-du) \]This reverses the limits of integration (when substituting \(-du\) accounts for this) and simplifies to:\[\int_{0}^{\pi} \frac{\pi u - 1}{\sqrt{u^2 + \pi^2}} \, du.\]This is identical to the original integral due to symmetry, indicating specific cancellations.

Analyze Integral Component-wise

Notice both transformed integrals end up with same form due to symmetry: \( \int_{0}^{\pi} \frac{\pi x - 1}{\sqrt{x^2 + \pi^2}} \, dx \) and its transformed equivalent. Each integral component involving symmetric properties such as the numerically opposing components added ultimately simplify at this bounded integral evaluation.

Evaluate Result based on Symmetry

Because of symmetry-based end transformations, our earlier steps deduce that each component within the integral offsetting leads to zero net area. Thus:\[\int_{0}^{\pi} \frac{\pi x - 1}{\sqrt{x^{2} + \pi^{2}}} \, dx = 0.\]

Key concepts

Integration Techniques

When facing an integral, identifying the correct technique is crucial for success. Integration encompasses various approaches designed to tackle different types of functions. Knowing which technique to use can streamline solving complex issues.
Integration techniques include:

• Substitution: Involves changing variables to simplify the integral.
• Integration by parts: Utilizes the product rule of differentiation in reverse to handle products of functions.
• Partial fractions: Useful for breaking down rational functions into simpler components.
• Trigonometric identities: Helps manipulate integrals involving trigonometric functions.

In the given exercise, the integrand presents a rational function with a denominator square root, hinting at substitution or utilizing symmetry. Recognizing these features plays an instrumental role in determining the most effective path forward.

Variable Substitution

Variable substitution is a powerful technique often employed to transform the integrand into a simpler form. It involves changing variables to a new variable (frequently denoted as \(u\)) that reconfigures the expression, rendering it more manageable.
In the original exercise, the substitution \(u = \pi - x\) was used. This transformed the limits and sign of the integral:

• The differential changes from \(dx\) to \(-du\), affecting the integration limits.
• The limits of integration reverse: from \(x = 0\) to \(x = \pi\) becomes \(u = \pi\) to \(u = 0\).

Thus, the integral, initially daunting, reshapes into the same expression with a variable swap. By maneuvering the expression through substitution, we're often able to draw insights or find simplification pathways that weren't evident initially.

Symmetry in Integrals

Symmetry plays a crucial role when evaluating integrals, especially over specific bounds. An integral over a symmetric interval often simplifies evaluation when the function exhibits symmetrical properties.
In this example, the interval spans from \(0\) to \(\pi\), and the function shows a symmetry that transforms upon substitution. With the substitution \(u=\pi-x\), the integral's form remains unchanged:

• The integral becomes, once transformed, mirror-like in its limits and maintains the same functional structure.
• This directly points out the inherent symmetry within the components and allows us to conclude the net area under the curve totals zero.

Such evaluations highlight why checking symmetry in functions and intervals is vital; they provide elegant insights and often lead directly to resolution without complicated calculation.