Problem 94 - Precalculus

Question

$$\int_{S \pi / 3}^{11 \pi / 6} \csc x \cot x d x$$

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Identify the Integral

The given integral is \( \int_{S \pi / 3}^{11 \pi / 6} \csc x \cot x \, dx \). We have to find the antiderivative of \( \csc x \cot x \) before evaluating it within the limits.

Step 2: Find the Antiderivative

Recall the derivative of \( \csc x \) which is \( -\csc x \cot x \). Thus, the antiderivative of \( \csc x \cot x \) is \( -\csc x \).

Step 3: Apply the Fundamental Theorem of Calculus

Using the antiderivative we found, apply the Fundamental Theorem of Calculus: \[ \int_{S \pi / 3}^{11 \pi / 6} \csc x \cot x \, dx = \left[ -\csc x \right]_{S \pi / 3}^{11 \pi / 6} \] Calculate the expression at the upper and lower limits.

Step 4: Evaluate at Upper Limit

Evaluate \( -\csc x \) at the upper limit \( x = \frac{11\pi}{6} \): \[ -\csc\left(\frac{11\pi}{6}\right) = -\frac{1}{\sin(\frac{11\pi}{6})} = -2 \] since \( \sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2} \).

Step 5: Evaluate at Lower Limit

Evaluate \( -\csc x \) at the lower limit \( x = \frac{S\pi}{3} \). Assume \( S \) is 5, giving \( \frac{5\pi}{3} \): \[ -\csc\left(\frac{5\pi}{3}\right) = -\frac{1}{\sin(\frac{5\pi}{3})} = -2 \] since \( \sin\left(\frac{5\pi}{3}\right) = -\frac{1}{2} \).

Step 6: Calculate the Integral

Finally, subtract the values and find the integral: \[ -2 - (-2) = 0 \] Thus, the value of the integral is 0.

Key Concepts

Antiderivative

An antiderivative, sometimes referred to as an indefinite integral, is a fundamental concept in calculus. It functions as the reverse of derivation. While the derivative of a function represents its rate of change, an antiderivative essentially rebuilds the original function from its derivative.
For the function \( \csc x \cot x \), recognizing its antiderivative is crucial to solving definite integrals. To find the antiderivative, you might remember from derivatives that the derivative of \( \csc x \) is \( -\csc x \cot x \). Hence, \( -\csc x \) acts as the antiderivative for \( \csc x \cot x \). This knowledge allows us to transition smoothly into evaluating definite integrals.
Understanding antiderivatives is not just about memorization, but about linking functions to their derivatives. This insight aids in solving more elaborate calculus problems by recognizing patterns and employing logic.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a bridge between two central operations in calculus: differentiation and integration. It consists of two main parts that establish the relationship between antiderivatives and definite integrals.
The first part tells us that if we differentiate an antiderivative, we retrieve the original function. The second part is more relevant here: It states that we can evaluate a definite integral by taking the difference between the values of an antiderivative evaluated at the upper and lower bounds.
For the integral \( \int_{S\pi / 3}^{11\pi / 6} \csc x \cot x \, dx \), once we find the antiderivative \(-\csc x\), applying the theorem involves evaluating \([-\csc x]_{S\pi / 3}^{11\pi / 6}\). In doing so, we find the difference between the values of \(-\csc x\) at \(x = \frac{11\pi}{6}\) and \(x = \frac{S\pi}{3}\). This gives us the definite integral's value, making the theorem a powerful tool in both practical and theoretical mathematics.

Trigonometric Functions

Trigonometric functions, like \( \csc x \) and \( \cot x \), are vital in calculus due to their recurring presence in various mathematical problems. They originate from right-angle triangles, with functions like sine and cosine forming the basis for others, such as cosecant (\( \csc x = 1/\sin x \)) and cotangent (\( \cot x = \cos x / \sin x \)).
Understanding these functions' behaviors and properties is crucial, especially when dealing with integrals in calculus. For instance, the derivative and antiderivative relationships are notable: knowing that the derivative of \( \csc x \) results in \( -\csc x \cot x \) helps us quickly find antiderivatives needed for integration.
Further, recognizing specific trigonometric angles and their sine values, such as \( x = \frac{11\pi}{6} \) where \( \sin\left( \frac{11\pi}{6} \right) = -\frac{1}{2} \), is essential for accurately evaluating integrals. Grasping these trigonometric tools enhances your ability to solve complex calculus problems effectively.