Problem 66 - Calculus for Biology and Medicine

Question

Compute the indefinite integrals. $$ \int \cos \frac{2-4 x}{5} d x $$

Step-by-Step Solution

Answer

The integral is \(-\frac{5}{4} \sin \left( \frac{2-4x}{5} \right) + C\).

Step 1: Simplify the Expression

Rewrite the integral in a simplified form. The integral given is \( \int \cos \left( \frac{2-4x}{5} \right) \, dx \). By letting \( u = \frac{2 - 4x}{5} \), evaluate \( du \) in terms of \( dx \).

Step 2: Substitute Variables

Calculate the derivative of \( u \). We have \( u = \frac{2 - 4x}{5} \), so by differentiating, \( du = \frac{-4}{5} \, dx \) or \( dx = \frac{5}{-4} \, du \). Substitute these into the integral.

Step 3: Integrate with Respect to New Variable

Substitute \( u \) and \( du \) into the integral: \( \int \cos(u) \cdot \frac{5}{-4} \, du \). This simplifies to \( -\frac{5}{4} \int \cos(u) \, du \). Calculate this indefinite integral.

Step 4: Find the Integral

Recall that \( \int \cos(u) \, du = \sin(u) + C \), where \( C \) is the constant of integration. Thus, the integral becomes \( -\frac{5}{4} \sin(u) + C \).

Step 5: Substitute Back the Original Variable

Substitute back \( u = \frac{2-4x}{5} \) into the expression. Thus, the original integral becomes \( -\frac{5}{4} \sin \left( \frac{2-4x}{5} \right) + C \).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique to calculate difficult integrals. It simplifies the process by changing variables to make the integration easier. When we handle an integral of a composite function, often in a format where direct integration isn't straightforward, this method comes in handy.

Here's how it works:

• Identify a part of the integral to substitute and set it as a new variable. This simplifies the expression. In this exercise, we set \( u = \frac{2 - 4x}{5} \).
• Differentiate this substitution to find the derivative, \( du \), in terms of the original variable, \( dx \). For our integral, \( du = \frac{-4}{5} \, dx \).
• Express \( dx \) in terms of \( du \) by rearranging the derivative equation. Here, \( dx = \frac{5}{-4} \, du \).
• Replace all parts of the integral involving \( x \) with your expressions for \( u \) and \( du \). This makes the integral simpler: \( \int \cos(u) \cdot \frac{5}{-4} \, du \).

Integration becomes easier after substitution and often results in a standard integral we know how to solve.

Trigonometric Integrals

The integral we worked on initially involved a trigonometric function, \( \cos \). Trigonometric integrals involve the integration of trigonometric functions like sine, cosine, tangent, and their variations.

In this integral, once we simplified it through substitution, we ended up with the integral of a single trigonometric function, \( \int \cos(u) \, du \). Solving this integral is straightforward because:

• The integral of \( \cos(u) \) is a standard form and is equal to \( \sin(u) + C \).
• We apply the integration constant \( C \) since the result is an indefinite integral.
• Trigonometric integrals can vary in complexity, but often substitution helps simplify them into forms that can be directly integrated using standard facts. Always check if substitution could simplify your work.

Understanding and recognizing these standard results makes dealing with trigonometric integrals much more manageable.

Constant of Integration

When dealing with indefinite integrals, the constant of integration, denoted by \( C \), is one essential component we can't ignore.

The constant of integration arises because integration is the reverse process of differentiation, and a function's derivative loses its constant term. This means:

• When we integrate a function without limits, infinitely many antiderivatives exist, each differing by a constant. Thus, we represent this family of solutions with \( + C \).
• In our solution, after integrating \( \int \cos(u) \, du \), we obtained \( \sin(u) + C \).
• The constant ensures all possible initial conditions from a differential equation (or scenario) are covered. It comes from the principle that the derivative of constant is zero, leaving room for any such missing term initially.

Understanding the relevance of \( C \) is critical for defining the complete behavior and family of functions that satisfy the original integration problem.