Problem 72 - Calculus Early Transcendentals: Pearson New International Edition

Question

. Evaluate the integral \(\int \cot x \csc ^{2} x d x\) by parts in two different ways: (a) By differentiating \(\cot x\) (b) By differentiating \(\csc x\) (c) Show that the two results are equivalent up to a constant.

Step-by-Step Solution

Answer

Both methods yield integrals differing by a constant.

Step 1: Set up Integration by Parts Formula

The integration by parts formula is given by \( \int u \, dv = uv - \int v \, du \). To solve the integral \( \int \cot x \, \csc^2 x \, dx \), we need to choose \( u \) and \( dv \) appropriately.

Step 2a: Differentiating \( \cot x \)

Let \( u = \cot x \) and \( dv = \csc^2 x \, dx \). Then, differentiate \( \cot x \) to get \( du = -\csc^2 x \, dx \). Integrate \( \csc^2 x \, dx \) to get \( v = -\cot x \).

Step 3a: Apply Integration by Parts Formula

Substitute into the formula: \( \int \cot x \, \csc^2 x \, dx = -\cot x \cdot \cot x - \int -\cot x \, (-\csc^2 x \, dx) \). Simplify to obtain \( -\cot^2 x - \int \cot x \, \csc^2 x \, dx \).

Step 4a: Solve the Resulting Equation

Identify and move the integral to one side: \( \int \cot x \, \csc^2 x \, dx + \int \cot x \, \csc^2 x \, dx = -\cot^2 x \). This simplifies to \( 2\int \cot x \, \csc^2 x \, dx = -\cot^2 x \). Finally, solve for the integral to find \( \int \cot x \, \csc^2 x \, dx = -\frac{1}{2}\cot^2 x + C \).

Step 2b: Differentiating \( \csc x \)

Let \( u = \csc x \) and \( dv = \cot x \, \csc x \, dx \). Differentiate \( \csc x \) to get \( du = -\csc x \, \cot x \, dx \). Integrate \( \cot x \, \csc x \, dx \) to get \( v = -\csc x \).

Step 3b: Apply Integration by Parts Formula

Substitute into the formula: \( \int \csc x \, (\cot x \, \csc x) \, dx = -\csc x \cdot \csc x - \int -\csc x \, (-\csc x \, \cot x \, dx) \). Simplify to obtain \( -\csc^2 x - \int \csc^2 x \, \cot x \, dx \).

Step 4b: Simplify and Solve

Simplify and identify duplicate steps: \( \int \csc^2 x \, \cot x \, dx + \int \csc^2 x \, \cot x \, dx = -\csc^2 x \). Simplify to \( 2\int \csc^2 x \, \cot x \, dx = -\csc^2 x \) and solve to find \( \int \csc^2 x \, \cot x \, dx = -\frac{1}{2}\csc^2 x + C \).

Step 5: Compare Results

The integrals \( \int \cot x \, \csc^2 x \, dx = -\frac{1}{2}\cot^2 x + C_1 \) and \( \int \csc^2 x \, \cot x \, dx = -\frac{1}{2}\csc^2 x + C_2 \) differ only by a constant, since \( \cot^2 x \) and \( \csc^2 x \) are related through the identity \( \csc^2 x = 1 + \cot^2 x \).

Key Concepts

Trigonometric Integrals

When dealing with trigonometric integrals, especially those that involve functions like cotangent (\( \cot x \)) and cosecant (\( \csc x \)), it is important to have a well-rounded understanding of identities and relationships between these functions. Integration involving trigonometric functions often requires the use of identities to simplify the expressions or to find a suitable substitution.

• For example, the identity \( \csc^2 x = 1 + \cot^2 x \) can be quite useful.


• This identity is instrumental in expressing integrals in different forms and simplifying the integration process.

Trigonometric integrals may also necessitate applying techniques such as substitution or integration by parts, as these integrate these functions more efficiently. In the given exercise, we see how \( \int \cot x \csc^2 x \, dx \) is handled using integration by parts, both by differentiating \( \cot x \) and \( \csc x \).

Calculus Techniques

Calculus equips us with powerful techniques to handle complex integrals. The integration by parts technique is based on the product rule for differentiation and is expressed as: \( \int u \, dv = uv - \int v \, du \). This method is particularly useful when dealing with products of functions where traditional methods of integration aren't easily applicable.
In this exercise, choosing which function to differentiate makes a significant difference. Differentiating \( \cot x \) versus differentiating \( \csc x \) shows how calculus techniques can yield equivalent results when approached from different angles:

• Step 2a involves letting \( u = \cot x \), which gives \( du = -\csc^2 x \, dx \) after differentiation.


• Alternatively, step 2b lets \( u = \csc x \), resulting in \( du = -\csc x \cot x \, dx \).

The choice of how to apply integration by parts reflects its flexibility and demonstrates the importance of understanding the function relationships and their derivatives.

Integral Equivalence

Integral equivalence occurs when different methods of integration result in the same function plus or minus a constant. In the exercise, both integration paths produce results that are equivalent up to a constant. This happens due to the trigonometric identity \( \csc^2 x = 1 + \cot^2 x \).

• Finding \( \int \cot x \csc^2 x \, dx = -\frac{1}{2}\cot^2 x + C_1 \) when differentiating \( \cot x \) and \( \int \csc^2 x \cot x \, dx = -\frac{1}{2}\csc^2 x + C_2 \) when differentiating \( \csc x \).
• The results differ by the constant arising from substituting the trigonometric identity \( \csc^2 x = 1 + \cot^2 x \). This allows the conversion from one form to the other.

Understanding integral equivalence is essential for verifying solutions and ensuring that methods are correctly applied in calculations. Recognizing these relationships helps in simplifying results and ensuring correctness.