Problem 30 - Calculus for Biology and Medicine

Question

In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int \cos (2 x-1) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \sin(2x - 1) + C \).

Step 1: Identify the Substitution

We want to use substitution to evaluate the integral \( \int \cos (2x - 1) \, dx \). Let's set \( u = 2x - 1 \). This simplifies the expression inside the cosine.

Step 2: Compute the Differential

Differentiate \( u = 2x - 1 \) with respect to \( x \) to find the differential \( du/dx = 2 \). Rearrange to find \( dx \) in terms of \( du \): \( du = 2 \, dx \) or \( dx = \frac{1}{2} \, du \).

Step 3: Substitute in the Integral

Substitute \( u \) and \( dx \) in the integral: \[ \int \cos(u) \, \frac{1}{2} \, du \]. This simplifies the integral to \( \frac{1}{2} \int \cos(u) \, du \).

Step 4: Integrate with Respect to u

The integral \( \int \cos(u) \, du \) results in \( \sin(u) + C \), where \( C \) is the constant of integration. Therefore, the integral becomes \( \frac{1}{2} \sin(u) + C \).

Step 5: Back-Substitute to the Original Variable

Replace \( u \) with the original expression in terms of \( x \): \( u = 2x - 1 \). Thus, the solution in terms of \( x \) is \( \frac{1}{2} \sin(2x - 1) + C \).

Key Concepts

Indefinite Integral

Indefinite integrals are a way to find antiderivatives of functions. When you integrate a function, you are essentially finding a function whose derivative gives you the original function. This is the reverse of differentiation. There's no need to specify bounds in an indefinite integral because it represents a family of functions, each differing by a constant. The general form for an indefinite integral is

• \( \int f(x) \, dx = F(x) + C \)
• where \( F(x) \) is the antiderivative of \( f(x) \)
• \( C \) is the constant of integration.

The goal is to find \( F(x) \) such that \( F'(x) = f(x) \). To solve an indefinite integral, use techniques like substitution or recognize standard integral forms. Remember, since the derivative of a constant is zero, any constant may appear in the solution.

Substitution Method

The substitution method is a valuable technique to simplify integrals when direct integration isn't straightforward. It essentially replaces part of the integral with a new variable, eliminating complexity. The key steps include:

• Identify part of the integral as a function \( u(x) \), typically within a composite function.
• Express \( dx \) in terms of \( du \); commonly, you find \( du = u'(x) \, dx \).
• Transform the integral using \( u \) and \( du \), which simplifies computation.
• Integrate with respect to \( u \).
• Back-substitute \( u \) to express the result in terms of the original variable.

For example, if faced with \( \int \cos(2x - 1) \, dx \), setting \( u = 2x - 1 \) simplifies the expression. Differentiating, \( du = 2 \, dx \) leads to \( dx = \frac{1}{2} \, du \). Substituting these into the integral transforms it into a simpler form, \( \frac{1}{2} \int \cos(u) \, du \), which is more straightforward to compute.

Trigonometric Integrals

Trigonometric integrals involve integrals of trigonometric functions like sine, cosine, and tangent. These integrals can often be daunting because of their repetitive and oscillating nature but are quite manageable once familiar with certain patterns and rules.

• The integral of \( \cos(u) \) is \( \sin(u) + C \).
• The integral of \( \sin(u) \) is \( -\cos(u) + C \).
• Integrals for tangent, cotangent, secant, and cosecant have their respective antiderivative forms.

The trick to managing trigonometric integrals, particularly when substitution comes into play, involves recognizing the function type and appropriate substitution that makes the integral simpler. In our example integral \( \int \cos(2x - 1) \, dx \), after substitution, it reduces to \( \int \cos(u) \, du \), which integrates directly to \( \sin(u) + C \). Finally, replace \( u \) with \( 2x - 1 \) to return to the original variable. Learning these integral basics unravels much larger and complex problems.