Problem 24 - Calculus Single Variable

Question

Integrate by parts successively to evaluate the given indefinite integral. $$ \int x^{2} \sin (x) d x $$

Step-by-Step Solution

Answer

\( \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C \).

Step 1: Choose Functions for Integration by Parts

According to integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we need to choose \( u \) and \( dv \). For this integral, let \( u = x^2 \) and \( dv = \sin(x) \, dx \).

Step 2: Differentiate and Integrate

Differentiate \( u \) to get \( du \): \[ du = 2x \, dx \] Now integrate \( dv \) to get \( v \):\[ v = \int \sin(x) \, dx = -\cos(x) \]

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), and \( v \) into the integration by parts formula:\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) - \int -\cos(x) \cdot 2x \, dx \]Simplify this expression:\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2 \int x \cos(x) \, dx \]

Step 4: Second Integration by Parts

Apply integration by parts again to \( \int x \cos(x) \, dx \). Let \( u = x \) and \( dv = \cos(x) \, dx \). Then \( du = dx \) and:\[ v = \int \cos(x) \, dx = \sin(x) \]Now substitute into the integration by parts formula:\[ \int x \cos(x) \, dx = x \sin(x) - \int \sin(x) \, dx \]This simplifies to:\[ \int x \cos(x) \, dx = x \sin(x) + \cos(x) \]

Step 5: Substitute and Simplify

Substitute the result from the second integration by parts back into the expression obtained in Step 3:\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2(x \sin(x) + \cos(x)) \]Expand and rearrange:\[ \int x^2 \sin(x) \, dx = -x^2 \cos(x) + 2x \sin(x) + 2\cos(x) + C \]where \( C \) is the constant of integration.

Key Concepts

Indefinite Integrals

An indefinite integral is a fundamental concept in calculus that represents the collection of all antiderivatives of a function. It is denoted by the integral symbol \( \int \) and provides a way to reverse the process of differentiation. When we calculate an indefinite integral, our main goal is to find a function whose derivative matches the given function. In other words, we are determining the "original function" before differentiation.

For example, if we want to find the indefinite integral of \( \sin(x) \), we are seeking a function whose derivative is \( \sin(x) \). The answer is \( -\cos(x) \), so:

• \( \int \sin(x) \, dx = -\cos(x) + C \)

where \( C \) is the constant of integration, reflecting the infinite set of antiderivatives.In the case of \( \int x^2 \sin(x) \, dx \), our task is more complex, as it involves a product of two functions. This requires applying specific techniques such as integration by parts, especially when simpler methods like direct integration or substitution won't work.

Integration Techniques

Integration techniques are diverse strategies used to solve integrals that cannot be easily evaluated by basic methods. Among these techniques, integration by parts is a crucial method especially useful for integrals involving products of algebraic and transcendental functions.

The formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

This technique is inspired by the product rule for differentiation and allows us to transform a complex integral into a simpler one. The key step in using this method effectively is wisely choosing \( u \) and \( dv \):

• \( u \) should be chosen such that its derivative \( du \) simplifies the integral.
• \( dv \) should be selected so it can be easily integrated to find \( v \).

In our exercise, we applied integration by parts twice. First, by choosing \( u = x^2 \) and \( dv = \sin(x) \, dx \), then we differentiated \( u \) and integrated \( dv \) to substitute into the formula and simplify the expression. Later, we encountered another integral, \( \int x \cos(x) \, dx \), which called for another round of integration by parts to solve completely.

Calculus Tutorial

Delving into calculus, especially integral calculus, involves understanding how to handle various types of functions and expressions. A tutorial on solving indefinite integrals, as seen with integration by parts, focuses on grasping the underlying principles that govern these techniques. This includes learning when and how to employ different methods depending on the function complexity.

In a calculus tutorial, we'd emphasize the significance of:

• Breaking down complex integrals into manageable parts.
• Identifying suitable integration methods for specific problem classes.
• Practicing repeatedly with diverse examples to build intuition and skill in selecting the most efficient technique.

To truly master the integration by parts strategy used in \( \int x^2 \sin(x) \, dx \), students should practice:

• Selecting the right \( u \) and \( dv \)
• Carrying through the differentiation and integration steps accurately
• Substituting back carefully to simplify the result

Understanding each step in the process, rather than simply memorizing it, fosters a deeper appreciation and capability in handling calculus challenges effectively.