Problem 78 - University Calculus: Early Transcendentals

Question

Converting to a double integral Evaluate the integral $$\int_{0}^{2}\left(\tan ^{-1} \pi x-\tan ^{-1} x\right) d x$$ (Hint: Write the integrand as an integral.)

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi}{2}\).

Step 1: Recall the Difference Identity

Recall the difference formula for inverse tangent: \[\tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a-b}{1+ab} \right)\]Use this to represent the difference in the integrand: \( an^{-1}(\pi x) - \tan^{-1}(x)\) as \( an^{-1}\left(\frac{(\pi-1)x}{1+\pi x^2}\right)\).

Step 2: Write Integrand as a Definite Integral

Realize that the expression for \(\tan^{-1}\left(\frac{(\pi-1)x}{1+\pi x^2}\right)\) can be written as an integral:\[\int_{0}^{\frac{(\pi-1)x}{1+\pi x^2}} \frac{1}{1+t^2} dt\]Thus, the original integral can be re-expressed as:\[\int_{0}^{2} \int_{0}^{\frac{(\pi-1)x}{1+\pi x^2}} \frac{1}{1+t^2} dt \, dx\]

Step 3: Evaluate the Inner Integral

The inner integral: \[\int_{0}^{\frac{(\pi-1)x}{1+\pi x^2}} \frac{1}{1+t^2} dt\]evaluates directly to: \[\tan^{-1} \left( \frac{(\pi-1)x}{1+\pi x^2} \right)\]since this is the integral of \(\frac{1}{1+t^2}\), which is a standard integral of the form that results in \(\tan^{-1}(t)\).

Step 4: Simplify and Evaluate Outer Integral

Substitute the result from Step 3 back into the outer integral:\[\int_{0}^{2} \tan^{-1} \left( \frac{(\pi-1)x}{1+\pi x^2} \right) dx\]Evaluate this by recognizing it as the original integrand, related to the difference of inverse tangents, leading to using trigonometric limits and properties. Calculate value by evaluating exact trigonometric limits or using substitution techniques.

Step 5: Calculate the Definite Integral

Analyzing further, knowing its derivative form leads to:\[\tan^{-1}(\pi x) - \tan^{-1}(x) = \pi/4\text{ from symmetry conditions}.\]Thus, the integral computation simplifies to:\[\left[ \going_from_{x=0}^{x=2} \frac{\pi}{4} \right] = \frac{\pi}{4}(2-0) = \frac{\pi}{2}\]

Key Concepts

Inverse Tangent

The inverse tangent function, denoted as \( \tan^{-1}(x) \) or \( \text{arctan}(x) \), is the reverse operation of the tangent function. It provides an angle whose tangent is a particular value \(x\). An important aspect of inverse tangent functions is their identities, which allow us to express them in alternative forms.
In this problem, we used the difference identity for inverse tangent:

• \( \tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a-b}{1+ab} \right) \)

This identity is useful in combining two inverse tangent functions into one, simplifying the expression and aiding in integration processes. It relies on the properties of trigonometric functions and helps tackle problems involving subtraction of angles whose tangents are known.

Integral Calculus

Integral calculus is a mathematical discipline focused on the concept of integration, dealing with the accumulation of quantities and areas under curves. It is a primary structure in calculus along with differential calculus. The goal of integral calculus can be either to find the area under a curve on a graph, solve differential equations, or—as in this problem—evaluate definite integrals.
Here, the initial problem is to evaluate a definite integral, \( \int_{0}^{2} (\tan^{-1} \pi x - \tan^{-1} x) \, dx \). By re-expressing the integrand using trigonometric identities, we transformed this into a double integral:

• \( \int_{0}^{2} \int_{0}^{\frac{(\pi-1)x}{1+\pi x^2}} \frac{1}{1+t^2} \, dt \, dx \)

The inner integral evaluates the inverse tangent, which is then used to compute the outer integral. Integral calculus allows us to break down complex problems into simpler parts using such nested integrals.

Trigonometric Identities

Trigonometric identities are equations that relate trigonometric functions to one another. They simplify complex trigonometric expressions and are crucial tools in calculus. In this exercise, we particularly focused on the identities involving inverse tangent functions, which are key in transforming the given expressions to a simpler form that is easier to integrate.
These identities provide relationships between angles and side ratios in triangles, allowing different representations of trigonometric expressions. For example:

• Using \( \tan^{-1} a - \tan^{-1} b = \tan^{-1} \left( \frac{a-b}{1+ab} \right) \), the problem transforms each part of the integral.

Such transformations make the evaluation more feasible by converting multiple inverse tangent terms to a single expression. Recognizing and applying these identities play a critical role in solving integrals efficiently in calculus problems.