Problem 24 - Calculus: Graphical, Numerical, Algebraic

Question

In Exercises \(21-24,\) use tabular integration to find the antiderivative. $$\int x^{3} \cos 2 x d x$$

Step-by-Step Solution

Answer

The antiderivative of the function is \(\frac{1}{2} x^{3} \sin 2 x - 3 x^{2}\frac{1}{2}\sin 2x + 6x\frac{1}{2}\cos 2x - 6\frac{1}{2}\sin 2x + C\), where C is the constant of integration.

Step 1: Selection of 'u' and 'dv'

In the integral \(\int x^3 \cos 2x \) dx, let 'u' be \(x^3\) and 'dv' be \(\cos 2x \) dx. This is because a simpler integral can be obtained if \(x^3\) is differentiated compared to \(\cos 2x\).

Step 2: Find 'du' and 'v'

Differentiate 'u' to find 'du', and integrate 'dv' to find 'v'. Therefore, 'du' is \(3x^2 dx\) and 'v' is \(\frac{1}{2} \sin 2x\).

Step 3: Apply Tabular Integration

Write a table with 'u' on the top and 'dv' at the bottom. The idea is to repeatedly differentiate 'u' and integrate 'dv' until 'u' becomes zero. 'u' becomes zero after three differentiations, so the process stops there. Then, alternate between plus and minus signs for each row. Start with a plus and end with a minus. Now, multiply the terms diagonally.

Step 4: Write the Answer

Add up the terms obtained from Step 3. The integral becomes \(\frac{1}{2} x^{3} \sin 2 x - 3 x^{2}\frac{1}{2}\sin 2x + 6x\frac{1}{2}\cos 2x - 6\frac{1}{2}\sin 2x\)

Key Concepts

Tabular Integration

Tabular integration is a streamlined method for applying integration by parts, particularly helpful when one function continuously differentiates to zero. This method is especially efficient for integrals of polynomial and trigonometric functions.

• Start by identifying suitable parts: one part to differentiate repeatedly and the other to integrate.
• Create a table with columns for 'u' functions, their derivatives, and 'dv' functions with their integrals.
• Alternate signs when multiplying diagonally, starting with a positive sign.
• Continue the process until you reach zero in the differentiation column.

Tabular integration simplifies calculations and reduces the possibility of errors in multiple integration by parts steps.

Integration by Parts

Integration by parts is a fundamental technique used to solve integrals by breaking them into simpler parts. It is particularly useful when the integrand is a product of functions.
The formula is derived from the product rule for differentiation: \[ \int u \, dv = uv - \int v \, du \]

• Choose 'u' such that its derivative is simpler, and the remaining part as 'dv'.
• Differentiate 'u' to find 'du'.
• Integrate 'dv' to find 'v'.

This method requires strategic choices for 'u' and 'dv' to simplify the resulting integrals, often integrating iteratively or using tabular integration.

Trigonometric Integrals

Trigonometric integrals involve trigonometric functions and often require specific techniques or substitutions. These may include sine, cosine, tangent, and other trigonometric functions.
When dealing with such integrals, consider the following:

• Function identities: Use identities like \( \sin^2 x + \cos^2 x = 1 \) to simplify expressions.
• Double angle formulas: These can transform the integral into a more manageable form.
• Use substitution: Simplify the integral using different trigonometric transformations when possible.

Trigonometric integrals often pair well with other techniques, such as integration by parts, to achieve solutions.

Differentiation

Differentiation is a core concept in calculus that involves finding the rate at which a function changes. It is essential for understanding integrals, particularly in finding antiderivatives.
Key concepts include:

• The derivative of a function is its slope or rate of change at any given point.
• Use rules like the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \) for polynomial functions.
• For trigonometric functions, know formulas such as \( \frac{d}{dx}(\sin x) = \cos x \).

Understanding differentiation is crucial when deconstructing integrals using integration by parts or tabular integration.