Problem 6 - Calculus for Biology and Medicine

Question

In Problems 1-30, use integration by parts to evaluate each integral. $$ \int x \sin (1-2 x) d x $$

Step-by-Step Solution

Answer

The integral \( \int x \sin(1-2x) \, dx = \frac{1}{2} x \cos(1-2x) + \frac{1}{4} \sin(1-2x) + C \).

Step 1: Identify Parts for Integration by Parts

In integration by parts, we use the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int x \sin(1-2x) \, dx \), choose \( u = x \) and \( dv = \sin(1-2x) \, dx \).

Step 2: Differentiate and Integrate the Parts

Differentiate \( u = x \) to get \( du = dx \). For \( dv = \sin(1-2x) \, dx \), we need to integrate to find \( v \). Let \( w = 1-2x \), then \( dw = -2 \, dx \). So, \( dx = -\frac{1}{2} \, dw \). The integral becomes \[ v = \int \sin(w) \left(-\frac{1}{2}\right) \, dw = -\frac{1}{2} (- \cos(w)) = \frac{1}{2} \cos(1-2x) \].

Step 3: Apply the Integration by Parts Formula

Now apply the formula: \( \int u \, dv = uv - \int v \, du \). Substitute \( u = x \), \( v = \frac{1}{2} \cos(1-2x) \), \( du = dx \):\[ \int x \sin(1-2x) \, dx = x \cdot \frac{1}{2} \cos(1-2x) - \int \frac{1}{2} \cos(1-2x) \, dx \].

Step 4: Simplify and Evaluate the Remaining Integral

The next integral is \( \int \frac{1}{2} \cos(1-2x) \, dx \). Again let \( w = 1-2x \), \( dw = -2 \, dx \), so \( dx = -\frac{1}{2} \, dw \). Thus, the integral becomes\[ \frac{1}{2} \int \cos(w) \, \left(-\frac{1}{2}\right) \, dw = -\frac{1}{4} \int \cos(w) \, dw = -\frac{1}{4} \sin(w) = -\frac{1}{4} \sin(1-2x) \].

Step 5: Combine Results and Simplify

Now, combine the terms:\[ \int x \sin(1-2x) \, dx = \frac{1}{2} x \cos(1-2x) + \frac{1}{4} \sin(1-2x) + C \]where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

A definite integral involves the evaluation of an integral over a specified interval from a to b. This results in a numerical value and represents the net area under the curve of the integrand between these two points. To denote this, we use the symbol \( \int_{a}^{b} f(x) \, dx \). When working with definite integrals, the rules and methods for indefinite integrals (with no upper and lower bounds) still apply, but the outcome is different. Instead of an expression involving a constant of integration, results contain actual values after applying the Fundamental Theorem of Calculus:

• Find the antiderivative, \( F(x) \), of the integrand.
• Evaluate \( F \) at the bounds: \( F(b) - F(a) \).

Understanding definite integrals is crucial since they help compute quantities such as areas, displacement, and total accumulation over time or space.

Trigonometric Integration

Trigonometric integration involves integrating expressions containing trigonometric functions, such as sine, cosine, and tangent. These require specific techniques or substitutions for simplification. In the context of integration by parts, as seen in our problem, knowing how to handle trigonometric functions is essential.For example, trigonometric identities like these can be useful:

• \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \)
• \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

Additionally, substituting portions of the functions, as done using \( w = 1 - 2x \) in the step-by-step solution, helps integrate with more straightforward trigonometric expressions. Simplifying trigonometric integrals can drastically reduce the complexity of calculus tasks.

Calculus Problem Solving

Instead of treating calculus problem-solving as procedural, it's beneficial to think of it as a strategic approach involving various techniques. Integration by parts, the method used here, is just one of many techniques. It involves recognizing parts within the integral that lend themselves to this formula.Typical steps to solving calculus problems include:

• Identify components of the integral (like \( u \) and \( dv \) for integration by parts).
• Differentiate and integrate appropriately (handling derivatives and antiderivatives).
• Simplify by applying algebraic operations and possibly other calculus techniques.

Each problem can be unique, requiring flexible thinking and combining different strategies. Visualizing the problem and understanding the mathematical properties involved are key in effective calculus problem-solving.