Problem 88 - Precalculus

Question

In calculus, the definite integral \(\int_{a}^{b} f(x) d x\) is used to find the area below the graph of a continuous function \(f\), above the \(x\) -axis, and between \(x=a\) and \(x=b\). The Fundamental Theorem of Calculus cstablishes that the definite integral \(f_{a}^{b} f(x) d x\) equals \(F(b)-F(a),\) where \(F\) is any antiderivative of a continuous function \(f.\) In Exercises \(87-90\), first shade the area corresponding to the definite integral and then use the information below to find the exact value of the area. $$\begin{array}{|l|c|c|c|c|} \hline \text { Function } & \tan x & \cot x & \sec x & \csc x \\ \hline \text { Antiderivative } & -\ln |\cos x| & \ln |\sin x| & \ln |\sec x+\tan x| & -\ln |\csc x+\cot x| \\ \hline \end{array}$$ $$\int_{\pi / 4}^{\pi / 2} \cot x d x$$

Step-by-Step Solution

Answer

The value of the integral is \(\frac{1}{2}\ln(2)\).

Step 1: Identify the Antiderivative

From the given table, the antiderivative of \ \(\cot x\) is \ \(\ln |\sin x|\). This matches the information we need to perform the integration.

Step 2: Apply the Fundamental Theorem of Calculus

Use the Fundamental Theorem of Calculus, which states \ \(\int_{a}^{b} f(x) dx = F(b) - F(a)\), where \ \(F\) is an antiderivative of \ \(f\{x}\). For our function, this becomes \ \(\int_{\pi/4}^{\pi/2} \cot x \, dx = \ln |\sin(\pi/2)| - \ln |\sin(\pi/4)|\).

Step 3: Evaluate the Antiderivative at the Upper Limit

Evaluate \ \(\ln |\sin(\pi/2)|\). Since \(\sin(\pi/2) = 1\), this becomes \ \(\ln(1) = 0\).

Step 4: Evaluate the Antiderivative at the Lower Limit

Evaluate \ \(\ln |\sin(\pi/4)|\). Since \(\sin(\pi/4) = \frac{\sqrt{2}}{2}\), this becomes \ \(\ln \left(\frac{\sqrt{2}}{2}\right)\).

Step 5: Calculate the Definite Integral

Substitute the values obtained in Steps 3 and 4 into the expression from Step 2: \ \(\ln |\sin(\pi/2)| - \ln |\sin(\pi/4)| = 0 - \ln \left(\frac{\sqrt{2}}{2}\right)\). This simplifies to \ \(\ln(2^{1/2}) = \frac{1}{2}\ln(2)\).

Step 6: Final Simplification

The expression \ \(\frac{1}{2}\ln(2)\) is the final value of the definite integral. If needed, this can be approximated using a calculator, but is often left in logarithmic form.

Key Concepts

Antiderivative

An antiderivative, also known as an indefinite integral, is a fundamental concept in calculus. It refers to a function whose derivative is the given function. Understanding this concept is crucial because finding antiderivatives is the first step in evaluating definite integrals.
For example, in the original problem, the function is \(\cot x\). By consulting the provided table, you can identify its antiderivative as \(\ln |\sin x|\). This is a vital step because without the antiderivative, you cannot proceed to calculate the definite integral.
Antiderivatives are not unique. If \(F(x)\) is an antiderivative of \(f(x)\), then \(F(x) + C\) is also an antiderivative, where \(C\) denotes a constant. In definite integration, however, this constant cancels out because it affects both end points of the evaluated integral equally.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus serves as a bridge between the concept of differentiation and integration, two of the primary operations in calculus.
This theorem states that if \(F\) is an antiderivative of a continuous function \(f\) on an interval \([a, b]\), then the definite integral of \(f\) from \(a\) to \(b\) is given by the change in \(F\) from \(a\) to \(b\):

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

In our solution, once the antiderivative \(\ln |\sin x|\) was found, we applied this theorem to evaluate the integral from \(\pi/4\) to \(\pi/2\).
Applying this, we calculated: \(\ln |\sin(\pi/2)| - \ln |\sin(\pi/4)|\). The result, as shown, simplifies to \(\frac{1}{2}\ln(2)\). This process underscores the powerful relationship between antiderivatives and definite integrals.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric expressions. These integrations often require recognizing patterns and identities that simplify the process.
In the given exercise, the integrand is \(\cot x\), a basic trigonometric function commonly encountered. \(\cot x\) is the quotient \(\frac{\cos x}{\sin x}\), and integrating such trigonometric functions sometimes needs the use of trigonometric identities or substitutions.

• For example, the derivative of \(\ln |\sin x|\) is \(\cot x\), making \(\ln |\sin x|\) the antiderivative of \(\cot x\).

By understanding and applying these identities, calculating the integral becomes a more straightforward task. Additionally, these concepts can simplify complex integrations involving multiple trigonometric terms.