Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \sin ^{5} 4 x \cos ^{2} 4 x d x\)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{4} \left( \frac{\cos^3(4x)}{3} - \frac{2\cos^5(4x)}{5} + \frac{\cos^7(4x)}{7} \right) + C\)."

Step 1: Recognize the Integral Type

The integral is of the form \( \int \sin^m(4x) \cos^n(4x) \, dx \) where \( m = 5 \) and \( n = 2 \). This suggests using trigonometric identities to simplify the integral.

Step 2: Use Trigonometric Identity

Use the identity \( \sin^2(u) = 1 - \cos^2(u) \) to rewrite \( \sin^5(4x) = \sin^4(4x) \cdot \sin(4x) = (1 - \cos^2(4x))^2 \cdot \sin(4x) \).

Step 3: Simplify the Expression

Expand \( (1 - \cos^2(4x))^2 \) to get \( 1 - 2\cos^2(4x) + \cos^4(4x) \). The integral becomes \( \int (1 - 2\cos^2(4x) + \cos^4(4x)) \cdot \sin(4x) \cdot \cos^2(4x) \, dx \).

Step 4: Integration by Substitution

Let \( u = \cos(4x) \), hence \( du = -4\sin(4x) \, dx \) or \( -\frac{1}{4} du = \sin(4x) \, dx \). Substitute these into the integral: \( -\frac{1}{4} \int (1 - 2u^2 + u^4) \cdot u^2 \, du \).

Step 5: Simplify the Substitute Integral

Now simplify to \( -\frac{1}{4} \int (u^2 - 2u^4 + u^6) \, du \).

Step 6: Integrate Term by Term

Integrate each term separately to get: \(-\frac{1}{4} \left( \frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7} \right) + C\).

Step 7: Substitute Back Original Variable

Replace \( u \) with \( \cos(4x) \):\(-\frac{1}{4} \left( \frac{(\cos^3(4x))}{3} - \frac{2(\cos^5(4x))}{5} + \frac{(\cos^7(4x))}{7} \right) + C\).

Key Concepts

Trigonometric Identities

Trigonometric identities are powerful tools that help simplify and manipulate trigonometric expressions. In the realm of integration, these identities can transform complex integrals into more manageable forms. One of the most commonly used identities is the Pythagorean identity, which states that \( \sin^2(u) + \cos^2(u) = 1 \).

• This identity allows us to express \( \sin^2(u) \) as \( 1 - \cos^2(u) \).
• Similarly, \( \cos^2(u) \) can be expressed as \( 1 - \sin^2(u) \).

In the provided solution, the identity \( \sin^2(u) = 1 - \cos^2(u) \) is used to transform the expression \( \sin^5(4x) \). By rewriting trigonometric powers in terms of a single trigonometric function, integrals become more approachable. This initial step is crucial for further simplification and integration techniques.

Integration by Substitution

Integration by substitution is a technique used to simplify integrals by changing variables. This process often involves identifying a part of the integral that can be substituted with a single variable. The goal is to transform the integrand into a simpler form. Here's how it works:

• Choose a substitution that simplifies the integral. For example, let \( u = \cos(4x) \).
• Differentiate your substitution: \( du = -4\sin(4x) \, dx \).
• Express \( dx \) in terms of \( du \) and the original variable: \( dx = -\frac{1}{4\sin(4x)} \, du \).

Substitution transforms the original integral into a new, possibly simpler problem. In the solution, the replacement \( u = \cos(4x) \) converts the trigonometric complexity into polynomial terms, which are easier to integrate. This technique is key in many integration problems, especially with trigonometric functions.

Power Reduction Formulas

Power reduction formulas are essential when dealing with trigonometric integrals that involve higher powers of sine or cosine. These formulas change the power of a trigonometric function into a sum of functions of lower powers or functions amenable to integration. For instance:

• Formulas like \( \cos^2(u) = \frac{1 + \cos(2u)}{2} \) help reduce high powers to simpler expressions.

In cases where powers are high, such as \( \sin^5(4x) \) or \( \cos^4(4x) \), initially rewriting these using power reduction formulas can greatly aid the simplification. This method effectively reduces the integral's complexity before engaging with other techniques such as substitution or integration by parts. Using these formulas correctly is a critical step in solving challenging integrals.