Problem 20 - Calculus

Question

Use the method of substitution to find each of the following indefinite integrals. $$ \int \cos (\pi v-\sqrt{7}) d v $$

Step-by-Step Solution

Answer

The integral is \( \frac{1}{\pi} \sin(\pi v - \sqrt{7}) + C \).

Step 1: Identify the Inner Function

Notice that the integral \( \int \cos(\pi v - \sqrt{7}) \, dv \) has an inner function, which is \( u = \pi v - \sqrt{7} \). This inner function will be used for substitution.

Step 2: Compute the Derivative of the Inner Function

Differentiate the inner function \( u = \pi v - \sqrt{7} \) with respect to \( v \). This gives \( \frac{du}{dv} = \pi \), so \( du = \pi \, dv \).

Step 3: Solve for dv

From the equation \( du = \pi \, dv \), solve for \( dv \): \( dv = \frac{du}{\pi} \).

Step 4: Substitute in Integral

Substitute \( u = \pi v - \sqrt{7} \) and \( dv = \frac{du}{\pi} \) into the original integral: \[\int \cos(u) \, \frac{du}{\pi} = \frac{1}{\pi} \int \cos(u) \, du.\]

Step 5: Integrate the Function

Integrate \( \frac{1}{\pi} \int \cos(u) \, du \). The integral is \( \sin(u) \), so \[\frac{1}{\pi} \sin(u) + C,\]where \( C \) is the constant of integration.

Step 6: Substitute Back the Original Variable

Replace \( u \) with the original expression: \[\frac{1}{\pi} \sin(\pi v - \sqrt{7}) + C.\]This is the final expression in terms of \( v \).

Key Concepts

Indefinite Integrals

Indefinite integrals are a fundamental concept in calculus. They represent the most general form of a function's antiderivative. When you perform an indefinite integration, you are essentially finding another function that, when differentiated, would yield the original function under the integral sign.
The result of an indefinite integral is not just a single function but a family of functions. This is because you add an arbitrary constant, denoted as \( C \), to indicate that an indefinite integral encompasses all possible antiderivatives.
The notation for indefinite integrals is "\( \int f(x) \, dx \)". Here, \( f(x) \) is the function you wish to integrate, and \( dx \) denotes the variable of integration. Indefinite integrals are essential for solving various calculus problems, especially when determining the accumulation of values or reversing the process of differentiation.

Calculus Problem-Solving

Solving calculus problems often involves a variety of techniques and strategies. It's essential to first understand the problem structure and the functions involved. Breaking the problem down into manageable steps is crucial to finding a solution.
For integration problems, especially, identifying whether you are dealing with an indefinite or definite integral is vital. Here are some general steps for calculus problem solving:

• Identify the type of integral (indefinite or definite).
• Analyze the integrand for any functions or expressions that can be simplified or require substitution.
• Determine appropriate methods for integration, be it substitution, integration by parts, or partial fractions.
• Perform the integration meticulously to avoid errors in algebraic manipulation.
• Check your work by differentiating your result to ensure it yields the original integrand.

Mastery of calculus problem-solving comes from practice and familiarity with a variety of functions and techniques.

Integration Techniques

Integration techniques are methods used to find the integral of a function. For calculus students, knowing these techniques is central to solving integration problems effectively.
One of the most useful techniques when dealing with complex functions is the substitution method, also known as "u-substitution." This method makes integrals easier to solve by simplifying the function under the integral sign.

• **Identify the Inner Function:** Pick a part of the integrand that you can define as a variable \( u \). This is often referred to as the inner function.
• **Differentiate the Inner Function:** Find the derivative of this new variable concerning the original variable.
• **Substitute and Simplify:** Substitute \( u \) and express \( dx \) or \( dv \) using \( du \). This change of variables should simplify the integrand.
• **Integrate and Back-Substitute:** Perform the integration with respect to \( u \), and then substitute back the original variable.

Substitution is particularly helpful for integrals with nested functions, like composite or trigonometric functions, making complex integration problems manageable and solvable.