Question

Different methods a. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\cot x\) b. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\csc x\) c. Reconcile the results in parts (a) and (b).

Short answer

In conclusion, we have shown that both substitution methods produce equivalent results for the integral \(\int \cot x \csc ^{2} x d x\). By using the identities and properties of trigonometric functions, we successfully reconciled the two solutions and demonstrated their equivalence. This exercise serves as a great example of how different methods in calculus can lead to the same correct result.

Step-by-step solution

Setup the Problem

We have the following two substitution methods: a. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\cot x\) b. Evaluate \(\int \cot x \csc ^{2} x d x\) using the substitution \(u=\csc x\)

Evaluate the Integral Using the First Substitution Method (u = cot x)

First, let's find the derivative of u with respect to x: \(\frac{d u}{d x} = \frac{d}{d x}(\cot{x}) = -\csc^2 x\) Now, we can rewrite the integral as: \(\int \cot x \csc ^{2} x d x = -\int u du\) Evaluating the integral: \(-\int u du = -\frac{1}{2}u^2 + C_1 = -\frac{1}{2}(\cot{x})^2 + C_1\)

Evaluate the Integral Using the Second Substitution Method (u = csc x)

First, let's find the derivative of u with respect to x: \(\frac{d u}{d x} = \frac{d}{d x}(\csc{x}) = -\csc x \cot x\) Now, we can rewrite the integral as: \(\int \cot x \csc ^{2} x d x = -\int \frac{du}{u}\) Evaluating the integral: \(-\int \frac{du}{u} = -\ln{|u|} + C_2 = -\ln{|\csc{x}|} + C_2\)

Reconcile the Results

To reconcile the results, we will first compare the two solutions and find a relationship between them. \(-\frac{1}{2}(\cot{x})^2 + C_1 = -\ln{|\csc{x}|} + C_2\) Let's use the identity \(\csc^2{x} - \cot^2{x} = 1\): Solving for \(\cot^2{x}\), we get: \(\cot^2{x} = \csc^2{x} -1\) Replace this in the first solution: \(-\frac{1}{2}(\csc^2{x} - 1) + C_1 = -\ln{|\csc{x}|} + C_2\) Now, let's write \(\ln{|\csc{x}|}\) using exponential form: \( |\csc{x}| = e^{C_2 - C_1 - \frac{1}{2}(\csc^2{x} - 1)}\) Since the left side is absolute value, we can see that the right side should be equal to either \(\csc{x}\) or \(-\csc{x}\). By observation, we can see that the positive case works. Therefore, we can say that, \(\csc{x} = e^{C_2 - C_1 - \frac{1}{2}(\csc^2{x} - 1)}\) Replacing this back in the second solution, we get \(-\ln{|\csc{x}|} + C_2 = -\ln{\left| e^{C_2 - C_1 - \frac{1}{2}(\csc^2{x} - 1)} \right|} + C_2\) Now, we use the property that \(\ln{e^x} = x\), so we have \(-\ln{|\csc{x}|} + C_2 = -\left( C_2 - C_1 - \frac{1}{2}( \csc^2{x} - 1) \right) + C_2\) Since the two solutions are equal, we have reconciled the results.

Key concepts

Integration by Substitution

Integration by substitution is a technique used to simplify the process of integrating complex functions. It involves changing the variable of integration to transform the integral into a simpler form. The basic idea is to choose a substitution that makes the integral easier to solve. In this exercise, we consider two substitution methods for integrating the same function, showcasing the flexibility of integration by substitution.

For example, when integrating \[\int \cot x \csc^2 x \, dx\]using the substitution \(u = \cot x\), we find that\[ \frac{du}{dx} = -\csc^2 x\] indicates that the original integral can be rewritten as \[-\int u \, du\]. This transforms the integral into a simple quadratic form:\[-\frac{1}{2}u^2 + C\]. By reverting back to the original \(x\) variable, we recognize this as \[-\frac{1}{2}(\cot x)^2 + C_1\].

Alternatively, by using the substitution \(u = \csc x\), \[\frac{du}{dx} = -\csc x \cot x\] allows the integral to be expressed as \[-\int \frac{du}{u}\]. Evaluating this results in \[-\ln{|u|} + C = -\ln{|\csc{x}|} + C_2\].

Both methods demonstrate the power of substitution in reducing complexity by choosing different variable transformations, highlighting multiple paths to arrive at the integral's solution.

Trigonometric Integrals

Trigonometric integrals involve the integration of functions that include trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. Such integrals are quite common and often require strategic manipulation to simplify the expression before integration.

In the current problem, are dealing with trigonometric functions, \(\cot x\) and \(\csc^2 x\)\,. Both functions are derivations of basic trigonometric identities that can make the integral complex in its original form. However, by examining and identifying appropriate trigonometric identities or derivatives, we can simplify the problem significantly.

For integration, understanding these trigonometric relationships is key. For example, we used derivative identities \(\frac{d}{dx}(\cot{x}) = -\csc^2 x\)\, and \(\frac{d}{dx}(\csc{x}) = -\csc x \cot x\), to determine what substitutions would be most beneficial. By efficiently applying these principles, we reduce difficulty and streamline solving these integrals. This reflects foundational skills in manipulating and utilizing trigonometric identities for integration purposes effectively.

Reconciliation of Integration Results

Reconciling integration results involves aligning outcomes from different integration methods to ensure consistency and correctness. This crucial skill verifies that, although alternative paths are taken, the fundamental solution aligns when reduced to its simplest equivalent form.

In the exercise, we have two solutions derived from separate substitution methods yielding different-looking outcomes:

• Using \(u = \cot x\): \\[-\frac{1}{2}(\cot{x})^2 + C_1\]
• Using \(u = \csc x\): \\[-\ln{|\csc{x}|} + C_2\]

Using the trigonometric identity \(\csc^2 x - \cot^2 x = 1\)\,, we rewrite \(\cot^2{x}\)\, such that it connects both results. Upon rearranging and re-substituting the relevant expressions, these solutions can be set to equivalency by aligning constant terms, affirming that both methods ultimately describe the same integrated form.

Achieving this requires recognizing equivalent expressions and the role constants of integration play in potentially transforming one result to match another. This underscores the importance of mathematical identities and the nuances behind managing constants in integration solutions.