Question

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem. $$ \int x^{2} \sin x d x $$

Short answer

Use integration by parts with \( u = x^2 \) and \( dv = \sin x \, dx \).

Step-by-step solution

Analyze the Integral

First, we need to analyze the given integral \( \int x^2 \sin x \, dx \). We have a polynomial \( x^2 \) and a trigonometric function \( \sin x \), which suggests a product of functions.

Choose a Strategy

When integrating a product of functions where one is a polynomial, integration by parts is typically a suitable technique. This is because we can differentiate the polynomial until it disappears.

Recall Integration by Parts Formula

Recall the formula for integration by parts: \( \int u \, dv = uv - \int v \, du \). We need to select parts of the integral that match the structure \( u \) and \( dv \).

Choose u and dv

For integration by parts, we usually set \( u \) to be the polynomial \( x^2 \), because the derivative will simplify the term. So, we choose \( u = x^2 \) and \( dv = \sin x \, dx \).

Verify Chosen u and dv

Differentiate \( u \) to get \( du = 2x \, dx \). Integrate \( dv \) to get \( v = -\cos x \). Since these derivations are straightforward, our choice seems appropriate for integration by parts.

Key concepts

Polynomial and Trigonometric Functions

Polynomials are mathematical expressions consisting of variables raised to whole number exponents. In an integral like \( \int x^2 \sin x \, dx \), the term \( x^2 \) represents a polynomial. Trigonometric functions like \( \sin x \) describe the relationships between the angles and sides of triangles and are periodic in nature. Both play significant roles in calculus due to their fundamental properties and applications in natural phenomena.

When these two functions are combined in an integral, such as in our problem, it indicates that a method of solving that effectively handles both expression types is required. Approaches like integration by parts are useful here as they leverage the structural simplicity of the polynomial when differentiated and the periodic nature of the trigonometric function when integrated. This facilitates breaking down the integral into manageable parts that can be more easily solved step by step.

Integration Techniques

Integration by parts is a powerful technique used to evaluate integrals involving products of functions, such as polynomials and trigonometric functions. It is based on the product rule for differentiation and is expressed using the formula:

\[ \int u \, dv = uv - \int v \, du \]

This formula highlights a strategic approach: choose part of the integrand as \( u \), which you will differentiate, simplifying it when possible. Choose the other part as \( dv \), which you will integrate. In the problem \( \int x^2 \sin x \, dx \), setting \( u = x^2 \) simplifies the polynomial through differentiation, while \( dv = \sin x \, dx \) integrates into \( v = -\cos x \).

Integration by parts shines in polynomial-trigonometric products, polynomials with exponentials, and logarithmic integrals, where differentiation reduces complexity or where integrations can be easily computed. Remember always to check if iterations of the formula are necessary, as some integrals, like ours, may need repeated applications.

Differentiation and Integration

Differentiation and integration are core operations in calculus, each serving distinct but complementary functions. Differentiation involves finding the derivative of a function, capturing its rate of change. For a polynomial like \( x^2 \), differentiating gives \( 2x \), greatly simplifying the expression in integration by parts.

Integration is essentially the reverse process, loosely described as finding the area under the curve of a function. While differentiation breaks functions down, integration often reconstructs them from their rates of change. When solving an integral using parts, you conceptually shift between these two processes.

For instance, our integral \( \int x^2 \sin x \, dx \) involved first differentiating \( u = x^2 \) to simplify into \( du = 2x \, dx \). We then integrate \( dv = \sin x \, dx \) to find \( v = -\cos x \). Together, these steps form an elegant synergy between differentiation and integration, allowing complex integral evaluation using a systematic approach.