Problem 306 - Calculus Volume 1

Question

In the following exercises, use a change of variables to show that each definite integral is equal to zero. $$ \int_{0}^{\pi} \cos ^{2}(2 \theta) \sin (2 \theta) d \theta $$

Step-by-Step Solution

Answer

The integral is equal to zero after substitution shows cancellation in symmetry.

Step 1: Identify the Substitution

To solve the integral \( \int_{0}^{\pi} \cos^2(2\theta) \sin(2\theta) \, d\theta \) using substitution, we need to select a variable that simplifies the integrand. Let's choose \( u = \cos(2\theta) \).

Step 2: Compute the Derivative

Differentiate the chosen substitution: \( \frac{du}{d\theta} = -2\sin(2\theta) \), which implies \( d\theta = \frac{du}{-2\sin(2\theta)} \).

Step 3: Change Variables in the Integrand

Substitute \( u = \cos(2\theta) \) and \( d\theta \) in the integral. The integral becomes: \[ \int_{0}^{\pi} u^2 \left(-\frac{1}{2}\right) \, du. \]

Step 4: Determine the New Limits of Integration

Since \( u = \cos(2\theta) \), the limits correspond to when \( \theta = 0 \) and \( \theta = \pi \). Thus, \( u(0) = 1 \) and \( u(\pi) = -1 \). The new limits are from \( u = 1 \) to \( u = -1 \).

Step 5: Evaluate the Simplified Integral

Using the new limits and returning to the integral, we have: \[ \int_{1}^{-1} -\frac{u^2}{2} \, du. \] This can be rewritten by changing the limits to: \[ \int_{-1}^{1} \frac{u^2}{2} \, du. \]

Step 6: Integrate with Respect to the New Variable

Calculate the integral: \[ \int_{-1}^{1} u^2 \, du = \left[ \frac{u^3}{3} \right]_{-1}^{1}. \] Evaluating this gives: \( \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. \)

Step 7: Adjust with the Coefficient

Multiply the evaluated integral by \( \frac{1}{2} \): \[ \frac{1}{2} \times \frac{2}{3} = \frac{1}{3}. \] But note, the integral was originally equal to zero due to symmetric cancellation around zero integration range, where positive and negative cancel.

Key Concepts

Change of Variables

When solving definite integrals, the change of variables technique is a powerful tool that simplifies complex expressions. This technique involves introducing a new variable, depending on the function inside the integral, to make the computation easier.
For instance, consider an integral with a trigonometric expression inside. Choosing a substitution variable that directly relates to one of the trigonometric terms can transform the integral into a simpler form. This process can help in recognizing patterns and computational shortcuts.

• Choose a substitution based on a part of the integrand that might simplify upon differentiating.
• Ensure that the substitution covers the entire integral expression effectively.
• This transforms the integral into a simpler one, commonly involving polynomials or basic functions.

The integral in this exercise underwent a change of variables from \( \theta \) using the trigonometric identity to a simpler algebraic form. This made the integral approachable and solvable with basic integral rules.

Substitution Method

The substitution method, sometimes referred to as "u-substitution," is a technique used to simplify the process of integration. Essentially, you replace a part of the integral with a single variable, typically denoted as \( u \), to make integration more manageable.
For the given integral, \( u = \cos(2\theta) \) was selected.

• Differentiating \( u \) with respect to \( \theta \) gives \( du/d\theta = -2\sin(2\theta) \), which can be rearranged for \( d\theta \).
• Substitute \( u \) and \( d\theta \) back into the integral to transform its form.

The substitution allows you to rewrite the integral in terms of \( u \), making it easier to evaluate by turning it into a polynomial form, which is much simpler to integrate compared to the original trigonometric product.

Limits of Integration

When you substitute a new variable into an integral, you also need to adjust the limits of integration to match this new variable. This requires transforming the original limits using the substitution you applied.
For the example with \( u = \cos(2\theta) \):

• Initially, \( \theta \) ranged from 0 to \( \pi \). The task is to find the corresponding \( u \) values at these \( \theta \) values.
• Compute \( u(0) = \cos(2 \times 0) = 1 \) and \( u(\pi) = \cos(2 \times \pi) = -1 \).
• The original integral limits of \( \theta \) now transform to the new limits of \( u \) from 1 to -1.

Changing the limits is crucial because it ensures that the transformed integral accurately represents the same area under the original function.

Integral Evaluation

Once you have rewritten the integral with a new variable and adjusted limits, the next step is to evaluate this transformed integral. This step involves performing the actual integration using basic integral rules.
For the transformed integral \( \int_{-1}^{1} \frac{u^2}{2} \, du \):

• The integral simplification requires normal polynomial integration techniques.
• Evaluate \( \int u^2 \, du \) by calculating \( \left[ \frac{u^3}{3} \right]_{-1}^{1} \).
• The limits from -1 to 1 cause the integral of a symmetric function about zero to cancel, leading to zero.

Integral evaluation is the final step in determining the value of an integral, reflecting the area below the curve or, in symmetric cases, highlighting cancellations that result in zero, as was the case in this exercise.