Question

In Exercises \(1-10,\) find the indefinite integral. $$\int x^{2} \cos \left(\frac{x}{2}\right) d x$$

Short answer

The indefinite integral of \(x^{2} \cos \left(\frac{x}{2}\right) dx\) is \(2x^{2}\sin \left(\frac{x}{2}\right) - 16x\cos\left(\frac{x}{2}\right) + 32\sin \left(\frac{x}{2}\right) + C\)

Step-by-step solution

Identifying u and dv

In order to apply the integration by parts, the integrand \(x^{2} \cos \left(\frac{x}{2}\right)\) needs to be decomposed to u and dv. Usually we choose u as a function which becomes simpler after differentiation and dv as a function which is easier to integrate. In our case, let \(u=x^{2}\) and \(dv=\cos \left(\frac{x}{2}\right) dx\).

Differentiate and Integrate

Differentiate u to get du and integrate dv to get v. In this case, \(du = 2x dx\) and \(v = 2\sin \left(\frac{x}{2}\right)\).

Apply Integration by Parts

We use the integration by parts formula: \(\int u dv = uv - \int v du\). Substituting the values of u, v, du and dv into the formula gives: \(x^{2} \cdot 2\sin \left(\frac{x}{2}\right) - \int 2\sin \left(\frac{x}{2}\right) \cdot 2x dx\).

Simplify and Integrate the Resulting Integral

Simplify the expression to: \(2x^{2}\sin \left(\frac{x}{2}\right) - 4 \int x\sin \left(\frac{x}{2}\right) dx\). Now use the method of integration by parts again for the remaining integral. This time let \(u=x\) and \(dv=\sin \left(\frac{x}{2}\right) dx\). Applying the formula of integration by parts and integrating will give \(-4x 4\cos\left(\frac{x}{2}\right) + 16 \int \cos \left(\frac{x}{2}\right) dx\). By integrating the last part, we get a final expression of \(-4x 4\cos\left(\frac{x}{2}\right) + 16 \cdot 2\sin \left(\frac{x}{2}\right) +C\).

Combine and Simplify the Result

Combining our previous step results, we have \(2x^{2}\sin \left(\frac{x}{2}\right) - 16x\cos\left(\frac{x}{2}\right) + 32\sin \left(\frac{x}{2}\right) + C\). This is the indefinite integral we were asked to find.

Key concepts

Integration by Parts

The technique of integration by parts is a powerful tool derived from the product rule for differentiation. It allows us to integrate products of two functions that may not be straightforward to integrate on their own.

According to the integration by parts formula, if you have an integral of the form \( \int u\, dv \), it can be transformed into \( uv - \int v\, du \), where \( u \) and \( dv \) are parts of the original integrand you choose. The crux of the method lies in differentiating \( u \) to get \( du \) and integrating \( dv \) to get \( v \).

Choosing \( u \) and \( dv \) appropriately is crucial. A good general rule is to let \( u \) be the part of the integrand that simplifies upon differentiation and \( dv \) be the part that does not get more complicated upon integration. The goal of integration by parts is essentially to transform an unwieldy integral into simpler parts that can ultimately be integrated.

U-Substitution

U-substitution is a method used to simplify integrals by substituting a part of the integrand with a new variable \( u \). This change of variables can simplify the integral into a more recognizable form, often allowing for straightforward integration.

The substitution usually looks for an 'inner function' within a composite function, and a differential part that fits with the derivative of this inner function. Once the substitution is made, the integral is rewritten in terms of \( u \) and its differential \( du \). After integration, we substitute back to the original variable to get the final answer.

U-substitution can be thought of as the reverse chain rule for integration. While the chain rule provides a way to differentiate composite functions, u-substitution provides a way to integrate them. Proper selection of \( u \) based on the intuition for the derivative of a likely inner function can simplify many integrals that at first glance may seem complex.

Differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate at which the function's value is changing at any given point. Essentially, it measures how a function's output value changes in response to changes in its input value.

In the context of the integration by parts, differentiation plays a critical role in the simplification of one part of the integrand. By finding the derivative of \( u \) to get \( du \) in the integration by parts formula, we transition from a product of functions to a simpler expression that we hope to integrate easily.

The derivative itself is foundational in calculus and offers a mathematical toolkit for analyzing and solving a myriad of problems in physics, engineering, economics, and beyond. Mastering differentiation not only aids in tackling integrals but also enhances one’s overall mathematical agility and problem-solving skills.

Integrate Trigonometric Functions

Integrating trigonometric functions often involves identifying patterns or applying specific techniques that simplify the integral. These functions, such as sine and cosine, frequently appear in integrands, both independently and in combination with other functions.

Special integration techniques, such as u-substitution, integration by parts, or even trigonometric identities, can be used to find the indefinite integrals of these trigonometric functions. In the given exercise, integration by parts was applied to the product of a polynomial and a trigonometric function, showcasing how trigonometric functions can be integrated using this method.

Understanding how to manipulate and integrate trigonometric functions is especially beneficial in fields that involve waveforms, oscillations, or circular motion, where these functions naturally arise.