Problem 45 - Calculus Early Transcendentals

Question

Use integration by parts to derive the following reduction formulas. $$\int x^{n} \cos a x d x=\frac{x^{n} \sin a x}{a}-\frac{n}{a} \int x^{n-1} \sin a x d x, \quad \text { for } a \neq 0$$

Step-by-Step Solution

Answer

Question: Using the derived formula, find the integral of x^2 * cos(ax) dx. Answer: To find the integral of x^2 * cos(ax) dx, we will use the derived reduction formula with n=2: $$\int x^{2} \cos a x d x = \frac{x^{2} \sin a x}{a} - \frac{2}{a} \int x^{1} \sin a x d x$$ Now, we need to solve the integral $\int x \sin a x d x$. We can use integration by parts again, this time choosing u(x) = x and v'(x) = \sin(ax). Therefore, u'(x) = 1 and v(x) = -(1/a) \cos(ax). Applying integration by parts to the integral, we get: $$\int x \sin a x d x = -\frac{x \cos a x}{a} - \int \left(-\frac{\cos a x}{a}\right) d x$$ Simplifying and integrating the remaining integral, we get: $$\int x \sin a x d x = -\frac{x \cos a x}{a} + \frac{1}{a^2} \int \cos a x d x$$ Integrating the cosine term, we have: $$\int x \sin a x d x = -\frac{x \cos a x}{a} + \frac{1}{a^2} \cdot \frac{1}{a} \sin a x$$ Now, substituting the evaluated integral for the remaining integral in the original derived formula, we get: $$\int x^{2} \cos a x d x = \frac{x^{2} \sin a x}{a} - \frac{2}{a} \left(-\frac{x \cos a x}{a} + \frac{1}{a^2} \cdot \frac{1}{a} \sin a x \right)$$ Simplifying the expression, we have: $$\int x^{2} \cos a x d x = \frac{x^{2} \sin a x}{a} + \frac{2x \cos a x}{a^2} - \frac{2\sin a x}{a^3} + C$$ where C is the constant of integration.

Step 1: Identify u(x) and v'(x) in the given integral

In the given integral, we will choose our functions, u(x) = x^n, and v'(x) =cos(ax). Therefore, we will have u'(x) = nx^(n-1) and v(x) = (1/a)sin(ax).

Step 2: Use the integration by parts formula on the given integral

Now, we will apply the integration by parts formula to the given integral using the functions we have defined: $$\int x^{n} \cos a x d x = \left(\frac{x^{n} \sin a x}{a}\right) - \int \left(\frac{n x^{n-1} \sin a x}{a}\right) d x$$

Step 3: Simplify and rewrite the formula

In the final integral, we can take the constant term (n/a) outside of the integral: $$\int x^{n} \cos a x d x = \frac{x^{n} \sin a x}{a} - \frac{n}{a} \int x^{n-1} \sin a x d x$$ Now, we have derived the reduction formula for the given integral using integration by parts. The final result is: $$\int x^{n} \cos a x d x=\frac{x^{n} \sin a x}{a}-\frac{n}{a} \int x^{n-1} \sin a x d x, \quad \text { for } a \neq 0$$

Key Concepts

Reduction Formula

Reduction formulas are an essential tool in the realm of calculus, especially when dealing with complex integrals that follow a pattern. The basic idea of a reduction formula is to express an integral in terms of another integral with a lower power. This simplification helps us solve intricate integrals by breaking them down into more manageable parts.

When you apply integration by parts to derive a reduction formula, you're essentially finding a way to lower the complexity of the function. For the integral \(\int x^n \cos(ax) \, dx\), the process begins by identifying parts of the integral that can be repeatedly reduced until a simpler solution is easily reached.

• Initially, choose appropriate functions for \(u(x)\) and \(v'(x)\).
• Apply integration by parts to obtain a simpler form of the original integral.

Reduction formulas are particularly useful in iterative processes where you have a series of similar integrals that need to be solved. Each step relies on the previous, thus forming a pattern or chain that leads to a comprehensive solution.

Calculus

Calculus is the branch of mathematics that studies how things change. At its core are two fundamental concepts: differential calculus and integral calculus. While differential calculus focuses on rates of change and slopes of curves, integral calculus is concerned with accumulation and the area under curves. Hopefully, this makes understanding the changes and accumulations in various contexts easier.

In the context of this exercise, we are delving into integral calculus, specifically using integration by parts. This method is a powerful tool when it comes to manipulating and evaluating the integral of a product of functions. In simple terms:

• Integration by parts allows you to turn a complex problem into a simpler one.
• It relies on a straightforward formula derived from the product rule of differentiation.

By using integration by parts efficiently, you can tackle multiple types of integrals, transform them into simpler forms, and find hidden connections between different aspects of a function.

Integral Calculus

Integral calculus is focused on finding the total size, value, or amount when you know how things are accumulating. One of its main applications is computing areas under curves, but the beauty of integral calculus is in its versatility. It allows us to solve problems across diverse scientific fields.

A core technique within integral calculus is integration by parts, particularly useful when faced with an integral of products, as in \(\int x^n \cos(ax) \, dx\). This technique links the challenging integral to a simpler form, effectively reducing the complexity step by step.

• Helps to solve integrals that are difficult to handle directly.
• Transforms the given problem into a manageable format.

Integration by parts is an indispensable part of the toolbox for students and professionals tackling integral calculus. When you master these techniques, you unlock the potential to solve practical problems involving areas, volumes, central points, and many other scenarios.