Problem 54 - Calculus: Early Transcendentals

Question

Evaluate the indefinite integral as an infinite series. $$ \int x^2 \sin (x^2) dx $$

Step-by-Step Solution

Answer

The integral evaluates to \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(2n+1)!(4n+3)} + C \).

Step 1: Substitute for Easier Integration

In this step, we perform a substitution to simplify the integral. Set \( u = x^2 \), which implies that \( du = 2x \,dx \). Therefore, \( x \,dx = \frac{1}{2} \, du \). This transforms the integral: \( \int x^2 \sin(x^2) \,dx = \int u \sin(u) \cdot \frac{1}{2} \, du \). This becomes \( \frac{1}{2} \int u \sin(u) \, du \).

Step 2: Integrate by Parts

Use integration by parts, where \( \int u \, dv = uv - \int v \, du \). Let \( v' = \sin(u) \) and \( v = -\cos(u) \), and \( u = u \) so \( du = 1 \, du \). Applying integration by parts: \( \frac{1}{2} (u \cdot (-\cos(u)) - \int (-\cos(u)) \, du) = \frac{1}{2} (-u \cos(u) + \int \cos(u) \, du) \).

Step 3: Solve the Remaining Simple Integral

Evaluate the integral \( \int \cos(u) \, du \), which is simple: \( \int \cos(u) \, du = \sin(u) \). Substitute this back to get: \( \frac{1}{2} (-u \cos(u) + \sin(u)) \).

Step 4: Substitute Back the Original Variable

Since \( u = x^2 \), substitute back: \( \frac{1}{2} (-(x^2) \cos(x^2) + \sin(x^2)) \).

Step 5: Express as an Infinite Series

Using the Maclaurin series expansion for \( \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \), express \( \sin(x^2) \) as \( \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)!} \). Consequently, \((x^2)\sin(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n+1)!} \).

Step 6: Integrate the Series Term-by-Term

Integrate the series term-by-term: \( \int x^2 \sin(x^2) \, dx = \int \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n+1)!} \right)dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int x^{4n+2} \, dx \).

Step 7: Compute the Indefinite Integral for Each Term

The integral of each term \( x^{4n+2} \) is \( \frac{x^{4n+3}}{4n+3} \). Therefore, \( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!(4n+3)} x^{4n+3} + C \), where \( C \) is the constant of integration.

Key Concepts

Integration by Parts

Integration by parts is a technique in calculus used to integrate products of functions. The formula for integration by parts is derived from the product rule for differentiation, and it is given by: \[ \int u \, dv = uv - \int v \, du \] To apply this rule, you must identify parts of the integral as either \( u \) or \( dv \). Then, you differentiate \( u \) to get \( du \), and integrate \( dv \) to get \( v \). The goal is to transform the original integral into a simpler one. In our problem, \( u \) was chosen as \( u \) and \( dv \) as \( \sin(u) du \). The transformed integral became manageable, allowing us to solve it step by step. Remember, choosing \( u \) and \( dv \) effectively can make complicated integrals more approachable.

Maclaurin Series

The Maclaurin series is a type of Taylor series expansion of a function around 0. It expresses functions as infinite sums of their derivatives at a single point. The Maclaurin series for a function \( f(x) \) is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(0)}{n!} x^n \] In our example, we utilized the Maclaurin series to express \( \sin(x^2) \). The series expansion for \( \sin(x) \) is: \[ \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] By substituting \( x^2 \) into the series, we obtained an expression that helped us write \( \sin(x^2) \) as an infinite series. This method is particularly useful for handling complex trigonometric integrals.

Substitution Method

The substitution method is a powerful tool for solving integrals. It involves changing variables to simplify the integral, making it easier to solve. In this exercise, by letting \( u = x^2 \), we transformed the original integral into one involving \( \sin(u) \). The derivative, \( du = 2x \, dx \), helped us replace \( x \, dx \) with \( \frac{1}{2} \) du, simplifying the process. The substitution method is useful in many cases where direct integration is difficult. It allows you to tackle the integral from a different perspective, often transforming it into a more familiar form.

Series Expansion

Series expansion allows functions to be expressed as infinite sums. This is particularly helpful in calculus to approximate functions and solve complex problems. A series expansion is typically constructed by adding an infinite number of terms from a function's derivatives. This technique facilitates the integration of functions that are otherwise complex when expressed in their standard form. In practical terms, this means you can take something like \( x^2 \sin(x^2) \) and transform it into a series where each term can be integrated separately. By focusing on each component, especially in more advanced calculus problems, these infinite sums provide a clear pathway to arrive at an integral's solution.