Problem 42 - Calculus Single Variable

Question

Evaluate the given definite integral by finding an antiderivative of the integrand and applying Theorem \(3 .\) $$ \int_{\pi / 6}^{\pi / 4} \csc (x) \cot (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2 - \sqrt{2} \).

Step 1: Understand the Integral

The given integral is a definite integral from \( \pi/6 \) to \( \pi/4 \) of the function \( \csc(x) \cot(x) \). Our task is to evaluate this by using the antiderivative of the function.

Step 2: Recall Relevant Antiderivative

Recall that the antiderivative of \( \csc(x) \cot(x) \) is \( -\csc(x) \). This means that the integral of \( \csc(x) \cot(x) \) with respect to \( x \) is \( -\csc(x) + C \), where \( C \) is the constant of integration.

Step 3: Apply The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \( F(x) \) is an antiderivative of \( f(x) \), then \( \int_a^b f(x) \ dx = F(b) - F(a) \). In this case, \( F(x) = -\csc(x) \), \( a = \pi/6 \), and \( b = \pi/4 \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( x = \pi/4 \) into the antiderivative: \( F(\pi/4) = -\csc(\pi/4) = -\frac{1}{\sin(\pi/4)} = -\frac{1}{\frac{\sqrt{2}}{2}} = -\sqrt{2} \).

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute \( x = \pi/6 \) into the antiderivative: \( F(\pi/6) = -\csc(\pi/6) = -\frac{1}{\sin(\pi/6)} = -\frac{1}{\frac{1}{2}} = -2 \).

Step 6: Compute the Definite Integral

Subtract the result of the lower limit evaluation from the upper limit evaluation: \( -\sqrt{2} - (-2) = -\sqrt{2} + 2 \).

Key Concepts

Antiderivative

In the realm of calculus, an antiderivative is a function whose derivative is the given function. Essentially, if you have a function \( f(x) \), an antiderivative \( F(x) \) would fulfill the condition that \( F'(x) = f(x) \). This concept is critical because it helps us find the integral of a function, which is essentially the "reverse" of differentiation.

The antiderivative represents an entire family of functions, often expressed as \( F(x) + C \), where \( C \) is the constant of integration. This constant is crucial as it indicates that there are infinitely many functions that can have the same derivative, differing by a constant. When dealing with definite integrals, however, this constant cancels out, which means we can focus on specific values without worrying about \( C \).

For this specific problem involving \( \csc(x) \cot(x) \), the antiderivative is \( -\csc(x) \). Therefore, when you find the integral \( \int \csc(x) \cot(x) \ dx \), it equals \( -\csc(x) + C \). Understanding how to determine and apply antiderivatives is pivotal for solving definite integrals using the Fundamental Theorem of Calculus.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges two important concepts in calculus: differentiation and integration. It has two primary parts that reveal how these seemingly opposite operations are intricately connected.

• The first part of the theorem asserts that if you have a continuous function \( f(x) \) over an interval \([a, b]\), and if \( F(x) \) is its antiderivative, then the integral of \( f(x) \) from \( a \) to \( b \) is given by \( F(b) - F(a) \).
• The second part states that the derivative of the integral of a function is the original function itself, essentially stating that differentiation and integration are inverse processes.

For our problem, the Fundamental Theorem of Calculus is employed to compute the definite integral \( \int_{\pi/6}^{\pi/4} \csc(x) \cot(x) dx \) by determining the antiderivative \( -\csc(x) \), and then evaluating it with the bounds \( a = \pi/6 \) and \( b = \pi/4 \). By calculating \( F(b) - F(a) \), i.e., \( -\csc(\pi/4) - (-\csc(\pi/6)) \), we discover the integral's value, linking these core ideas.

Trigonometric Integrals

Trigonometric integrals take center stage when functions involve trigonometric identities. These integrals often require knowledge of specific antiderivatives and manipulation using trigonometric formulas.

In this particular exercise, we integrate the function \( \csc(x) \cot(x) \), which is a product of two fundamental trigonometric functions. Understanding these components:

• The cosecant function, \( \csc(x) \), is the reciprocal of \( \sin(x) \).
• The cotangent function, \( \cot(x) \), is the reciprocal of \( \tan(x) \) and can also be written as \( \frac{\cos(x)}{\sin(x)} \).

To evaluate integrals involving these and other trigonometric functions, one needs to remember the basic identities and relationships. Because integrals can be complex, recognizing antiderivatives of such combinations is crucial. The known antiderivative of \( \csc(x) \cot(x) \) simplifies finding the definite integral. Tools from trigonometry are essential in reducing a given integral to a form that can be easily solved using calculus techniques, such as substitution or parts, although not always necessary in straightforward cases like this.