Problem 56 - Calculus Single Variable

Question

Evaluate the given integral by converting the integrand to an expression in sines and cosines. $$ \int 5 \sin ^{5}(x / 3) \cot ^{2}(x / 3) d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{15}{4} \sin^4 \left( \frac{x}{3} \right) - \frac{5}{2} \sin^6 \left( \frac{x}{3} \right) + C. \)

Step 1: Convert Cotangent to Sines and Cosines

The cotangent function can be expressed in terms of sines and cosines as \[ \cot \left( \frac{x}{3} \right) = \frac{\cos \left( \frac{x}{3} \right)}{\sin \left( \frac{x}{3} \right)}. \]Thus, the square of the cotangent function is \[ \cot^2 \left( \frac{x}{3} \right) = \left( \frac{\cos \left( \frac{x}{3} \right)}{\sin \left( \frac{x}{3} \right)} \right)^2 = \frac{\cos^2 \left( \frac{x}{3} \right)}{\sin^2 \left( \frac{x}{3} \right)}. \]

Step 2: Substitute Back into the Integral

Substituting the cotangent expression into the integral, we get:\[ \int 5 \sin^5 \left( \frac{x}{3} \right) \frac{\cos^2 \left( \frac{x}{3} \right)}{\sin^2 \left( \frac{x}{3} \right)} \, dx. \]Simplifying this expression, the integrand becomes:\[ 5 \sin^3 \left( \frac{x}{3} \right) \cos^2 \left( \frac{x}{3} \right). \]

Step 3: Apply a Simplified Trigonometric Identity

Utilize the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) to help later parts of the problem.Notice that using this identity can sometimes help express \( \cos^2(\theta) = 1 - \sin^2(\theta) \), which will be useful for further simplification.

Step 4: Use Substitution to Simplify Integration

Let \( u = \sin \left( \frac{x}{3} \right) \). Thus, \( du = \frac{1}{3} \cos \left( \frac{x}{3} \right) \, dx \) or rearranging it as\[ 3 \, du = \cos \left( \frac{x}{3} \right) \, dx. \]

Step 5: Express New Integral with 'u' Substitution

With the substitution, the integral becomes:\[ 5 \int u^3 (1 - u^2) 3 \, du. \]Distributing and simplifying this expression gives:\[ 15 \int (u^3 - u^5) \, du. \]

Step 6: Integrate the Polynomial Expression

Integrating term by term, we have:\[ 15 \left( \frac{u^4}{4} - \frac{u^6}{6} \right) + C. \]

Step 7: Substitute Back for Original Variable

Substitute \( u = \sin \left( \frac{x}{3} \right) \) back into the integral to get:\[ \frac{15}{4} \sin^4 \left( \frac{x}{3} \right) - \frac{15}{6} \sin^6 \left( \frac{x}{3} \right) + C. \]Finally simplifying gives:\[ \frac{15}{4} \sin^4 \left( \frac{x}{3} \right) - \frac{5}{2} \sin^6 \left( \frac{x}{3} \right) + C. \]

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial in simplifying complex trigonometric expressions. This skill helps in breaking down and transforming integrals for easier computation. In the given exercise, we start by expressing \( \cot \left( \frac{x}{3} \right) \) in terms of sine and cosine as \( \frac{\cos \left( \frac{x}{3} \right)}{\sin \left( \frac{x}{3} \right)} \). Trigonometric identities are like the toolkit that allows us to express trigonometric functions in different forms.
For example:

• The identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) is often used to interchange between \( \sin \) and \( \cos \).
• By using the identity \( \cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)} \), it becomes simpler to manipulate expressions that involve ratios of trigonometric functions.

By translating cotangent into sines and cosines, we simplify the integration process and prepare for further manipulation using substitution.

U-Substitution

U-substitution is a technique used to simplify the process of integration, particularly when dealing with composite functions. In this exercise, we use u-substitution by letting \( u = \sin \left( \frac{x}{3} \right) \). This choice is strategic because it aligns with the trigonometric substitution already in the equation.
Here's how u-substitution works:

• Identify the inner function to set as \( u \). In this case, \( u = \sin \left( \frac{x}{3} \right) \).
• Determine \( du \), which is the differential of \( u \). With our substitution, \( du = \frac{1}{3} \cos \left( \frac{x}{3} \right) \, dx \).
• Rearrange the equation to express \( dx \) in terms of \( du \) as \( 3 \, du = \cos \left( \frac{x}{3} \right) \, dx \).

Once the substitution is made, it transforms the integral into a polynomial form, making it straightforward to integrate.

Polynomial Integration

Once we have performed the u-substitution, the integral takes on a polynomial form: \( 15 \int (u^3 - u^5) \, du \). Polynomial integration is straightforward, as it involves applying the power rule for integration. The power rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \).
Here's how we integrate the polynomial:

• For \( u^3 \), apply the power rule to get \( \frac{u^4}{4} \).
• For \( u^5 \), apply the power rule to get \( \frac{u^6}{6} \).
• Combine these results: \( 15 \left( \frac{u^4}{4} - \frac{u^6}{6} \right) + C \).

Polynomial integration is quite methodical compared to other techniques since each term is dealt with separately, greatly simplifying our task.

Trigonometric Substitution

Trigonometric substitution is a powerful method that replaces complex expressions involving \( \sin, \cos, \) and \( \tan \) with simpler algebraic forms. In this exercise, the entire approach hinges on substituting trigonometric expressions into polynomial form for easier integration.
After substituting back the original trigonometric expression, \( u = \sin \left( \frac{x}{3} \right) \), we obtain our final integrated result in terms of the original variables.

• Re-substituting gives \( \frac{15}{4} \sin^4 \left( \frac{x}{3} \right) - \frac{5}{2} \sin^6 \left( \frac{x}{3} \right) + C \).
• This step is critical as it completes the cycle of substitution, returning the primitive function to its original trigonometric context.

This showcases the elegance of trigonometric substitution — by briefly transforming the trigonometric form into a polynomial, we leverage both algebra and trigonometry to find integrals effectively.