Problem 5 - Calculus for Biology and Medicine

Question

Use integration by parts to evaluate the integrals. $$ \int 2 x \sin (x-1) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(-2x \cos(x-1) + 2 \sin(x-1) + C\).

Step 1: Identify Parts for Integration by Parts

To apply integration by parts, identify a function for differentiation (u) and another function for integration (dv). For the integral \(\int 2x \sin(x-1) \, dx\), choose \(u = 2x\) and \(dv = \sin(x-1) \, dx\).

Step 2: Differentiate u and Integrate dv

Differentiate \(u\) to find \(du\) and integrate \(dv\) to find \(v\). We have \(du = 2 \, dx\). To find \(v\), integrate \(\sin(x-1)\), which gives \(v = -\cos(x-1)\).

Step 3: Apply Integration by Parts Formula

Use the integration by parts formula: \(\int u \, dv = uv - \int v \, du\). Substituting the values, we get:\[-2x \cos(x-1) - \int -\cos(x-1) \, 2 \, dx\].

Step 4: Simplify and Integrate Remaining Integral

Simplify the expression to:\[-2x \cos(x-1) + 2 \int \cos(x-1) \, dx\].The remaining integral is \(2 \int \cos(x-1) \, dx\), which evaluates to: \(2 \sin(x-1)\).

Step 5: Combine Solutions and Add Constant

Combine the results to obtain:\[-2x \cos(x-1) + 2 \sin(x-1) + C\], where \(C\) is the integration constant.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus used to find the derivative of a function. A derivative represents the rate of change of a function's value with respect to a change in its independent variable. In our exercise, we first differentiate the chosen part of the function, which is essential for applying integration by parts effectively.
When differentiating, we look for how a function behaves as its inputs change.

• This involves systematic rules, like the power rule, product rule, and chain rule.
• Derivatives provide critical information about graph behaviors, such as slopes and curves' concavity.

In practice, the derivative of a function is often expressed as \( du \).
For instance, in our example, we choose \( u = 2x \), and differentiate it to obtain \( du = 2 \, dx \). Differentiation here prepares this portion of the integral for further processing, helping us break down complex integrals into simpler parts.

Integration

Integration is the reverse process of differentiation, often referred to as finding the "antiderivative." It’s used to calculate the area under the curve represented by a function.
This process is central to solving integral-related problems, such as those you encounter when employing techniques like integration by parts.
In our example, integration is performed on selected function parts to facilitate solving the overall integral.

• Choose a segment of your function to be \( dv \).
• Use standard integration techniques to compute the antiderivative, resulting in \( v \).

For the integral \( \int 2x \sin(x-1) \, dx \), we choose \( dv = \sin(x-1) \, dx \), and integrate to find \( v = -\cos(x-1) \). Integration blends seamlessly with differentiation in this technique to break and solve the problem in parts, ultimately leading to a complete solution.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that are products or compositions with trigonometric functions like \( \sin \), \( \cos \), and \( \tan \). These types of integrals can be complex due to periodic variations of trigonometric functions.
Mastering the integration of such functions often entails recognizing patterns and identities inherent in trigonometry, simplifying the process.

• Common trigonometric identities, such as \( \sin^2 x + \cos^2 x = 1 \) or double angle formulas, can ease integration.
• Such expressions often need specific strategies, including integration by parts, substitution, or trigonometric identities, to simplify solutions.

In our exercise, the function \( \sin(x-1) \) involved a shift as part of its argument, yet it's managed through straightforward integration. By recognizing and handling these subtleties, we successfully derive the antiderivative \( v = -\cos(x-1) \). Successfully integrating or differentiating with trigonometric components enhances one’s calculus-solving skills, making future encounters with such equations far more manageable.