Problem 63 - Calculus: Early Transcendental Functions

Question

For \(a>0,\) show that \(\int_{a}^{1} \frac{1}{x^{2}+1} d x=\int_{1}^{1 / a} \frac{1}{x^{2}+1} d x .\) Use this equality to derive an identity involving tan \(^{-1} \bar{x}\)

Step-by-Step Solution

Answer

The identity involving the tan inverse function derived from the given integral equality is \(\tan^{-1}(a) = \tan^{-1}(1/a)\).

Step 1: Substitute \(1/x\) by \(u\)

Begin by letting \(u = 1/x\), and hence \(du = -dx/x^{2}\). This will transform the integral.

Step 2: Replace \(x\) with \(u\) in the integral

Replacing \(x\) by \(u\) in the integral, we have: \(\int_{1}^{1 / a} \frac{1}{x^{2}+1} d x = - \int_{1}^{a} \frac{1}{u^{2}+1}d(u)\).

Step 3: Demonstrate the Integrals are Equal

Now, it is shown that the original integrals are the same: \(\int_{a}^{1} \frac{1}{x^{2}+1} dx = - \int_{1}^{a} \frac{1}{u^{2}+1} du = \int_{1}^{1/a} \frac{1}{u^{2}+1} du\).

Step 4: Derive the identity involving tan inverse

It is known that the integral of \(1/(1+x²)\) is tan inverse of \(x\). So, the equality means that \(\tan^{-1}(1) - \tan^{-1}(a) = \tan^{-1}(1/a) - \tan^{-1}(1)\). Simplifying it, we get \(\tan^{-1}(a) = \tan^{-1}(1/a)\), which is the required identity.

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on concepts of integration, which is a fundamental tool for solving problems involving areas, volumes, and accumulated quantities. Integration reverses the process of differentiation, operating as an anti-derivative. In this exercise, we deal with the definite integrals from a specific range.
The process of integration by substitution is akin to differentiation's chain rule. We typically use substitution to simplify integrals by changing the variable or the limits of integration. In our solution, we substituted \(u = 1/x\), to simplify the computation. This change of variable helps to transform the integrals so that they can be evaluated more straightforwardly.
When working with definite integrals, remember:

• Change both the integrand and the limits of integration during substitution.
• If the limits are expressed in terms of the original variable, convert them to the new variable when you substitute.

This ensures that the integral retains its meaning and answers the original problem accurately.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. They simplify calculations and analyses in trigonometry, allowing us to transform complicated expressions into simpler ones.
In the context of integration problems, these identities often come in handy, particularly when dealing with functions like \( \frac{1}{x^2 + 1} \), which are related to the arctangent function. The identity used here is related to constructing known integrals from inverse trigonometric functions, specifically the inverse tangent or arctangent.
A crucial identity involved in our discussion is:

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

Using this, we derive further conclusions that simplify proving the equality of definite integrals.

Inverse Trigonometric Functions

Inverse trigonometric functions are fundamental when switching between angles and trigonometric ratios. They reverse the operations of the usual trigonometric functions sine, cosine, and tangent. For instance, \( \tan^{-1}(x) \) returns the arc or angle whose tangent is \(x\).
These functions are critical in integral calculus, especially when evaluating integrals involving expressions like \( \frac{1}{x^2 + 1} \). The integral of \( \frac{1}{x^2 + 1} \) over any range yields \( \tan^{-1}(x) \) plus a constant.
**Important aspects of inverse trigonometric functions include:**

• Range and domain: For \( \tan^{-1}(x) \), the domain is all real numbers, and the range is \((-\pi/2, \pi/2)\).
• The derivation of formulae: Being the inverse of \(tan\), usage of \( \tan^{-1} \) assists in transforming integrals into more manageable solutions.

These characteristics allow us to effectively evaluate and manipulate expressions involving inverse trigonometric functions.