Problem 9 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int \cot (3-7 x) d x $$

Step-by-Step Solution

Answer

The integral is \( -\frac{1}{7} \ln |\sin(3 - 7x)| + C \).

Step 1: Identify the Substitution

We are given the integral \( \int \cot(3 - 7x) \, dx \). Notice that \(3 - 7x\) is a linear expression of \(x\). To simplify this, let's choose the substitution \(u = 3 - 7x\). Differentiating both sides with respect to \(x\) gives \(du = -7 \, dx\). So, \(dx = -\frac{1}{7} \, du\).

Step 2: Substitute and Rewrite the Integral

Substitute \(u = 3 - 7x\) and \(dx = -\frac{1}{7} \, du\) into the original integral. The integral becomes: \[ \int \cot(u) \, (-\frac{1}{7}) \, du = -\frac{1}{7} \int \cot(u) \, du \]

Step 3: Integrate Using Standard Integral Forms

We know from standard integral tables that \( \int \cot(u) \, du = \ln |\sin(u)| + C \). Substitute this back into our integral to get: \[ -\frac{1}{7} \ln |\sin(u)| + C \]

Step 4: Back-substitute to Original Variable

Replace \(u\) with \(3 - 7x\) using our original substitution \(u = 3 - 7x\). Thus, the solution becomes: \[ -\frac{1}{7} \ln |\sin(3 - 7x)| + C \]

Step 5: Final Result

Conclude that the integral evaluates to \[ \int \cot(3 - 7x) \, dx = -\frac{1}{7} \ln |\sin(3 - 7x)| + C \], where \(C\) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals allow us to calculate the exact area under a curve between two points on a graph. They are denoted by an integral sign with upper and lower limits. For example, when evaluating \[ \int_a^b f(x) \, dx, \]\(a\) and \(b\) are the limits. This gives us a numerical value representing the accumulated area.
One key aspect of definite integrals is the Fundamental Theorem of Calculus. It links the concept of integration with differentiation. It states that if \(F(x)\) is an antiderivative of \(f(x)\), then:\[\int_a^b f(x) \, dx = F(b) - F(a)\]This relationship helps in evaluating integrals efficiently, especially when dealing with continuous functions.
While definite integrals focus on an actual range, the exercise addresses indefinite integrals, which involve finding a universal expression for the antiderivative without specific limits. Regardless, understanding definite integrals is foundational since it leads to determining specific values when needed.

Trigonometric Integrals

Trigonometric integrals involve functions with sine, cosine, tangent, and other trigonometric terms. These integrals are common in calculus due to the periodic nature and symmetry properties of trig functions.
For the integral of \(\cot(u)\) as seen in the exercise, the solution employs:

• Using known integrals and identities
• Substitution techniques
• Rewriting terms in simpler forms

The integral of \(\cot(u)\) is given by the formula:\[\int \cot(u) \, du = \ln |\sin(u)| + C\]Understanding this form is crucial as it helps to solve similar types of trigonometric integrals easily.
Additionally, substitution plays a key role in simplifying these integrals by converting them into more manageable forms. Many trigonometric integrals may not be immediately obvious, thus recognizing patterns and standard forms is essential for success in tackling them.

Calculus Techniques

In calculus, a variety of techniques are used to solve integrals and derivatives. The substitution method is particularly useful when dealing with complex integrals.
Substitution transforms a difficult integral into an easier one. In the presented exercise, substitution was used to turn the integral of \(\cot(3 - 7x)\) into a simpler form by setting:

• \(u = 3 - 7x\)
• \(du = -7 \, dx\) or \(dx = -\frac{1}{7} \, du\)

This change of variables makes the integral more concise and easier to solve.
Once substitution is complete and the integral has been evaluated, back-substitution returns the expression to the original variables. This step ensures results are expressed in the context of the original problem.
Mastering substitution and other calculus techniques is invaluable. It allows for versatility in approaching a wide range of problems, ensuring success in calculus tasks.