Problem 67 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \sin (2 x-1) d x $$

Step-by-Step Solution

Answer

The integral \( \int \sin (2x-1) \, dx = -\frac{1}{2} \cos(2x-1) + C \).

Step 1: Identify the Integral Type

The given integral is \( \int \sin(2x-1) \, dx \). This is a standard integration problem involving a trigonometric function, specifically the sine function. We will use substitution method to solve it.

Step 2: Choose the Substitution

Let \( u = 2x - 1 \). Then the differential \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \). This substitution will simplify the integral.

Step 3: Substitute and Simplify the Integral

Substitute \( u = 2x - 1 \) and \( dx = \frac{1}{2} \, du \) in the integral. The integral becomes \( \int \sin(u) \, \frac{1}{2} \, du = \frac{1}{2} \int \sin(u) \, du \).

Step 4: Integrate with Respect to u

The integral \( \int \sin(u) \, du \) is known to be \( -\cos(u) + C \). Therefore, \( \frac{1}{2} \int \sin(u) \, du = \frac{1}{2} (-\cos(u) + C) = -\frac{1}{2} \cos(u) + C_1 \), where \( C_1 \) is a constant of integration.

Step 5: Resubstitute x Back into the Solution

Substitute \( u = 2x - 1 \) back into the expression. You get \( -\frac{1}{2} \cos(2x - 1) + C_1 \).

Key Concepts

Trigonometric Integration

Trigonometric integration is a powerful technique used to solve integrals involving trigonometric functions like sine, cosine, and others. These functions are prevalent in many calculus problems, especially those involving periodic phenomena. The main aim is to simplify these integrals, often by using identities or substitutions that reduce complexity. When dealing with functions such as \(\sin(ax+b)\), where direct integration is not straightforward, trigonometric identities or substitution methods make the integration process more manageable. Here are a few common approaches in trigonometric integration:

• Identifying and utilizing basic trigonometric identities.
• Using the substitution method to transform the integral into a simpler form.
• Knowing antiderivatives of basic trigonometric functions, such as \(\int \sin(u) \, du = -\cos(u) + C\).

Understanding the behavior of trigonometric functions and their derivatives is essential for solving integrals involving them. As seen in the given problem, substitution played a significant role in simplifying the integration of \( \sin(2x-1) \).

Substitution Method

The substitution method is a fundamental tool in calculus, especially useful for solving integrals that are not straightforward. The core idea is to change variables in a way that simplifies the integral. This method often turns a challenging integral into one that is much easier to handle. In our problem, we deal with \( \int \sin(2x-1) \, dx\). The substitution method involves these steps:

• Choose a substitution: Here, setting \( u = 2x - 1 \) simplifies the integrand.
• Compute the differential: This involves finding \( du = 2 \, dx \) therefore \( dx = \frac{1}{2} \, du \).
• Replace original variables: The integral becomes \( \frac{1}{2} \int \sin(u) \, du \), which is easier to solve.

This method is especially useful when the derivative of part of the integrand matches (or appears in the integrand). Practicing the substitution method can greatly enhance one's ability to tackle a variety of calculus problems, enabling the transformation of complex integrals into manageable forms.

Calculus Problem Solving

Solving calculus problems requires understanding a blend of techniques and strategies, particularly when integrating complex functions. Being adept at identifying and applying the right method can significantly simplify the problem-solving process. Here are some key strategies that enhance calculus problem-solving:

• Analyze the Problem: Identify the type of function and the most appropriate integration technique to use.
• Choose the Right Method: Whether it's substitution, integration by parts, or trigonometric identities, selecting the correct method is crucial.
• Practice: Regular practice with a variety of problems ensures proficiency and confidence in using different calculus techniques.

Effective problem solving in calculus involves knowing not just the steps to solve an integral but understanding why these steps work. This deeper understanding helps in selecting the best approach to solve integrals, whether they involve trigonometric functions, polynomial expressions, or other complex derivatives, as showcased in the integration of \( \sin(2x-1) \).