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Magazine Chapter 7: Problem 11
Find the following indefinite and definite integrals. \(\int_{0}^{\pi} \sin ^{5}(3 x) \cos (3 x) d x\)
Short Answer
Expert verified
The definite integral evaluates to 0.
Step by step solution
01
Identify the integral to solve
We need to evaluate the definite integral: \[ \int_{0}^{\pi} \sin^{5}(3x) \cos(3x) \, dx \] This involves a power of sine and a cosine factor in a trigonometric integral.
02
Use a substitution method
Substitute \( u = \sin(3x) \). Then, \( du = 3\cos(3x) \, dx \), or equivalently \( dx = \frac{du}{3\cos(3x)} \).Adjust the integral to reflect this substitution: \[ \int \sin^{5}(3x) \cos(3x) \, dx = \int u^5 \cdot \frac{du}{3} = \frac{1}{3} \int u^5 \, du \] We will evaluate this indefinite integral first.
03
Integrate using a power rule for indefinite integral
Calculate the indefinite integral: \[ \int u^5 \, du = \frac{u^6}{6} + C \] So, substituting back, we have: \[ \frac{1}{3} \cdot \frac{u^6}{6} = \frac{u^6}{18} + C \] Replace \( u \) with \( \sin(3x) \): \[ F(x) = \frac{(\sin(3x))^6}{18} + C \]
04
Evaluate the definite integral from 0 to \(\pi\)
Now, use the antiderivative \( F(x) \) to evaluate the definite integral from 0 to \( \pi \):\[ \int_{0}^{\pi} \sin^{5}(3x) \cos(3x) \, dx = \left[ \frac{(\sin(3x))^6}{18} \right]_{0}^{\pi} \]Calculate: \[ \frac{(\sin(3 \cdot \pi))^6}{18} - \frac{(\sin(3 \cdot 0))^6}{18} = \frac{0^6}{18} - \frac{0^6}{18} = 0 \]
05
Conclude with the result
The definite integral of the function over the interval from 0 to \( \pi \) is 0, as the evaluated expression at both bounds is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method is a powerful technique used to simplify integral problems, particularly those involving trigonometric functions. In our exercise, we aim to ease the integration process by changing variables.
- Identify Patterns: The problem includes a product of trigonometric functions. In this case, it's convenient to set a substitution where the derivative of the substitution appears in the integrand.
- Set the Substitution: Here, we choose to let \( u = \sin(3x) \). This means that \( du = 3\cos(3x) \, dx \). This step aligns part of the integrand with its differential form outside the integral.
- Adjusting the Integral: After substitution, the integral looks simpler: \( \int \sin^5(3x)\cos(3x) \, dx = \int u^5 \cdot \frac{du}{3} \). This transformation makes it a polynomial integral in terms of \( u \), which is much simpler to solve.
Power Rule Integration
The power rule for integration is a fundamental tool for solving polynomial integrals. When you have a simple power of a variable, the power rule provides a straightforward approach to find the indefinite integral.
- General Formula: For a function \( u^n \), the power rule states that \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \), where \( n eq -1 \).
- Application: In our simplified problem from substitution, we have \( \int u^5 \, du \). Applying the power rule, we get \( \frac{u^6}{6} + C \), simplifying the polynomial integration process.
- Back-Substitution: Once the indefinite integral is found, substitute back the original function of \( x \). Thus, the integral solution \( \frac{u^6}{18} + C \) when replacing \( u \) back becomes \( F(x) = \frac{(\sin(3x))^6}{18} + C \).
Definite Integral Calculation
Definite integrals calculate the area under a curve between two limits. In contrast to indefinite integrals which include an arbitrary constant \( C \), definite integrals evaluate to a specific number.
- Evaluating Bounds: Once an antiderivative is found, the definite integral is calculated by evaluating the antiderivative at the upper bound and subtracting its evaluation at the lower bound.
- Plug into Limits: For our exercise, we found the antiderivative \( F(x) = \frac{(\sin(3x))^6}{18} \). Evaluating it from \( 0 \) to \( \pi \) means computing \( \left[ \frac{(\sin(3\cdot\pi))^6}{18} - \frac{(\sin(3\cdot 0))^6}{18} \right] \).
- Outcome: Both \( \sin(3\pi) \) and \( \sin(0) \) are zero, thus \( \frac{0^6}{18} - \frac{0^6}{18} = 0 \). The area under the curve for this integral over the interval is zero, often indicating symmetry or cancellation over the interval.
