Problem 32 - Calculus: One and Several Variables

Question

Calculate. $$\int_{0}^{\pi / 4} \frac{\arctan x}{1+x^{2}} d x$$

Step-by-Step Solution

Answer

The value of the given definite integral is \(\frac{\pi^2}{32}\).

Step 1: Choose u and dv

We use integration by parts formula: \[\int u \, dv = uv - \int v \, du\] Let \(u = \arctan(x)\) and \(dv = \frac{1}{1+x^2} dx\). Step 2: Calculate du and v

Step 2: Find du and v

Differentiate \(u\) to find \(du\), and integrate \(dv\) to find \(v\): \(du = \frac{1}{1+x^2} dx\) \(v = \int\frac{1}{1+x^2}dx = \arctan(x) + C\) Step 3: Apply the integration by parts formula

Step 3: Substitute u, v, du, and dv in the formula

Substitute our u, v, du, and dv values back into the integration by parts formula: \[\int \frac{\arctan(x)}{1+x^2} dx = \left(\arctan(x)\right)\left(\arctan(x) + C\right) - \int \left(\arctan(x) + C\right)\frac{1}{1+x^2} dx\] Since we are taking a definite integral, the constant C will be cancelled out. So we can remove it: \[\int_{0}^{\pi/4} \frac{\arctan(x)}{1+x^2} dx = \left.\left(\arctan(x)\right)^2\right|_0^{\pi/4} - \int_{0}^{\pi/4} \left(\arctan(x)\right)\frac{1}{1+x^2} dx\] Step 4: Evaluate the definite integral

Step 4: Evaluate the integrals at given limits

First, we need to evaluate \((\arctan(x))^2\) at \(\frac{\pi}{4}\) and \(0\): \[\left(\arctan\left(\frac{\pi}{4}\right)\right)^2 - \left(\arctan(0)\right)^2 = \left(\frac{\pi}{4}\right)^2 - 0^2 = \frac{\pi^2}{16}\] The second term is the integral we started with, so we have the following remarkable identity for the integral: \[\int_{0}^{\pi/4} \frac{\arctan(x)}{1+x^{2}} d x = \frac{\pi^2}{16} - \int_{0}^{\pi/4} \frac{\arctan(x)}{1+x^{2}} d x\] Step 5: Solve for the integral

Step 5: Isolate the integral

To find the value of the integral, we can isolate it: \[2 \int_{0}^{\pi/4} \frac{\arctan(x)}{1+x^{2}} d x = \frac{\pi^2}{16}\] \[\int_{0}^{\pi/4} \frac{\arctan(x)}{1+x^{2}} d x = \frac{\pi^2}{32}\] Therefore, the value of the given definite integral is \(\frac{\pi^2}{32}\).

Key Concepts

Definite Integral Calculation

A definite integral calculation involves evaluating the area under a curve defined by a function, from one point to another on the x-axis. In the context of the exercise at hand, we calculate the definite integral of the function \frac{\arctan x}{1+x^{2}}\begin pseudo \dx\end pseudo from \(x = 0\) to \(x = \frac{\pi}{4}\). The power of definite integrals extends beyond just finding areas; they're vital in physics for calculating quantities like displacement, work, and probability distributions.

For definite integration, understanding the limits of integration (in this case, from 0 to \(\frac{\pi}{4}\)) is crucial. We're essentially summing infinitesimally small products of the function's value and a small change in x (\(\Delta x\)) across this interval. This sum gives us a precise numerical value representing the

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\arctan(x)\), are the inverse operations of trigonometric functions. They allow us to find an angle whose trigonometric function results in a given value. For instance, \(\arctan(x)\) helps us determine the angle whose tangent is x. These functions are commonly found in trigonometry, calculus, and geometry.

One interesting property of inverse trigonometric functions, particularly relevant to this calculus problem, is their differentiation. For example, the derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\), which comes in handy when solving integrals involving \(\arctan(x)\), as seen in this exercise. This relationship between the function and its derivative is why \(\arctan(x)\) is chosen as part of the integration by parts formula.

Arctan Integration

Arctan integration involves integrating the inverse tangent function, \(\arctan(x)\). This function often appears in calculus problems where a function's rate of change is proportional to a quantity inversely related to another, squared quantity plus one. That's a mouthful, but in simpler terms, it shows up when dealing with slopes that change based on the vertical distance from a certain line.

In solving integration problems with \(\arctan(x)\), it's handy to remember that its derivative is \(\frac{1}{1+x^2}\), which is a common integrand. As you'll see in this exercise, the integral of \(\arctan(x)\) plays a significant role and demonstrates a clever use of calculus techniques.

Calculus

Calculus is a branch of mathematics focused on change and motion. It's split into two main areas: differentiation, which deals with rates of change, and integration, which deals with accumulation of quantities. The problem provided involves integration, which can be thought of as the 'inverse' of differentiation: it helps us reconstruct quantities from their rates of change.

Integration by parts, a technique used in our exercise, is a crucial method in calculus used to integrate products of functions. It's based on the product rule for differentiation and is often employed when the function being integrated is the product of two functions that are not easily integrable on their own. The method requires careful selection of which part of the function should be differentiated (\(du\)) and which should be integrated (\(v\)), dramatically simplifying the problem.