Problem 30 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \csc \left(\frac{v-\pi}{2}\right) \cot \left(\frac{v-\pi}{2}\right) d v$$

Step-by-Step Solution

Answer

The integral evaluates to \[-2\csc\left( \frac{v-\pi}{2} \right) + C\].

Step 1: Recognize the Trigonometric Identity

The integrand contains \( \csc \left(\frac{v-\pi}{2}\right) \cot \left(\frac{v-\pi}{2}\right) \). Recall the derivative identity: \( \frac{d}{dx}[ \csc(x) ] = - \csc(x) \cot(x) \). This suggests that the integral can be solved using a substitution method.

Step 2: Perform the Substitution

Let \( u = \frac{v-\pi}{2} \). Then \( \frac{du}{dv} = \frac{1}{2} \), or \( du = \frac{1}{2} dv \). This implies \( dv = 2 du \). Substituting these into the integral gives:\[ \int \csc(u) \cot(u) \cdot 2 \, du = 2 \int \csc(u) \cot(u) \, du \]

Step 3: Integrate Using Known Antiderivative

Use the antiderivative \( \int \csc(u) \cot(u) \, du = -\csc(u) + C \). Therefore, the integral becomes:\[ 2 \times (-\csc(u) + C) = -2\csc(u) + C' \]where \( C' = 2C \) is an arbitrary constant.

Step 4: Back-Substitute for Original Variable

Replace \( u \) with \( \frac{v-\pi}{2} \) in the antiderivative:\[ -2\csc\left( \frac{v-\pi}{2} \right) + C' \]This is the solution to the original integral.

Key Concepts

Trigonometric Identities

Trigonometric identities are extremely useful tools in calculus, especially when dealing with integrals involving trigonometric functions like sine, cosine, tangent, cosecant, secant, and cotangent. In the given exercise, we encounter the expression \( \csc\left(\frac{v-\pi}{2}\right) \cot\left(\frac{v-\pi}{2}\right) \), which is a product involving the cosecant and cotangent functions.

Recall that trigonometric identities allow us to transform these functions for easier computation or integration. Specifically, the derivative of the cosecant function, \( \frac{d}{dx}[ \csc(x) ] = - \csc(x) \cot(x) \), is particularly handy here. This identity tells us that the derivative of \( \csc(u) \) with respect to \( u \) results in \( -\csc(u) \cot(u) \), which hints at the possibility of reversing this process through integration.

Understanding and recognizing these identities plays a crucial role in determining the correct integration techniques or potential substitutions to simplify and solve integrals effectively.

Substitution Method

The substitution method is a fundamental technique often used to simplify complex integrals by transforming them into more familiar or manageable forms. In this exercise, the substitution involves setting \( u = \frac{v-\pi}{2} \). This change of variable aims to simplify the integrand, transforming our original expression into something that is easier to integrate.

Here's how substitution works:

• First, define a new variable for substitution, here it's \( u = \frac{v-\pi}{2} \).
• Next, differentiate this expression with respect to the original variable \( v \), giving \( \frac{du}{dv} = \frac{1}{2} \), or equivalently \( dv = 2 \, du \).
• Substitute \( u \) and \( dv \) into the original integral, replacing \( dv \) with \( 2 \, du \) to maintain equality.

This conversion simplifies the problem significantly, allowing us to use known antiderivatives, such as \( \int \csc(u) \cot(u) \, du \).

After integrating, it's essential to back-substitute the original variable to complete the solution. In this exercise, you replace \( u \) with \( \frac{v-\pi}{2} \) to express the antiderivative in terms of \( v \), arriving at the final solution.

Antiderivatives

Finding the antiderivative, also known as the indefinite integral, is a key step in solving integrals in calculus. An antiderivative of a function is essentially the reverse process of differentiation. This means, given the derivative, finding the original function before differentiation occurred.

For our exercise, after applying substitution, the problem reduces to integrating \( \int \csc(u) \cot(u) \, du \). The antiderivative of this specific integral is known and given by \( -\csc(u) + C \), where \( C \) is the constant of integration.

• Recognize the antiderivative of the expression \( \csc(u) \cot(u) \), which is a common result often memorized due to frequent use in trigonometry.
• Multiplication by constants, like here with 2, should be factored into the antiderivative, resulting in \( -2\csc(u) + C' \).

The final step involves back-substituting the original variable by replacing \( u \) with \( \frac{v-\pi}{2} \), hence expressing the antiderivative in the context of the original integral.

Antiderivatives form the backbone of integration, providing the necessary tools to transition from rates of change back to functions themselves.