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Chapter 8: Problem 17

Evaluate the definite integrals (a) \(\int_{0}^{\pi} \sin 5 x \sin 6 x d x\) (b) \(\int_{0}^{\pi} \sin ^{2} 5 x d x\)

Short Answer

Expert verified
(a) 0 (b) \(\frac{\pi}{2}\)

Step by step solution

01

Decompose the Product of Sines

For part (a), use the product-to-sum identities to simplify the integration process. The identity for product-to-sum is given by: \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \). Apply this to \( \sin 5x \sin 6x \).
02

Apply Product-to-Sum Identity

Using the identity in Step 1, rewrite the integral: \[ \int_{0}^{\pi} \sin 5x \sin 6x \, dx = \frac{1}{2} \int_{0}^{\pi} [\cos(x) - \cos(11x)] \, dx. \]
03

Integrate Cosine Functions

Separate the integral into two parts: \[ \frac{1}{2} \left( \int_{0}^{\pi} \cos(x) \, dx - \int_{0}^{\pi} \cos(11x) \, dx \right). \]Integrate each part: \( \int \cos(x) \, dx = \sin(x) + C \) and \( \int \cos(11x) \, dx = \frac{1}{11} \sin(11x) + C \).
04

Evaluate Integrals at Limits

Plug in the integration limits for each integral and evaluate: \[ \frac{1}{2} \left( [\sin(x)]_{0}^{\pi} - \frac{1}{11}[\sin(11x)]_{0}^{\pi} \right) \]. Since \( \sin(\pi) = \sin(0) = 0 \), both expressions \( \sin(x) \) and \( \sin(11x) \) evaluate to zero at the bounds, so the entire expression simplifies to zero.
05

Writing the Identity for Square of Sine

For part (b), use the identity \( \sin^2(A) = \frac{1 - \cos(2A)}{2} \). Apply this identity to \( \sin^2(5x) \) to simplify the integral process.
06

Simplify the Integral Expression

Using the identity \[ \int_{0}^{\pi} \sin^2(5x) \, dx = \int_{0}^{\pi} \frac{1 - \cos(10x)}{2} \, dx. \] This can be split into two separate integrals: \[ \frac{1}{2} \left( \int_{0}^{\pi} 1 \, dx - \int_{0}^{\pi} \cos(10x) \, dx \right). \]
07

Integrate and Evaluate

Evaluate each integral: \( \int_{0}^{\pi} 1 \, dx = \pi \) and \( \int_{0}^{\pi} \cos(10x) \, dx = \frac{1}{10}[\sin(10x)]_{0}^{\pi} \) which evaluates to zero. Thus, the original integral becomes \[ \frac{1}{2} \left( \pi - 0 \right) = \frac{\pi}{2}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

trigonometric identities
Understanding trigonometric identities is crucial when dealing with integrals involving trigonometric functions. These identities help simplify expressions and make integration easier. For instance, the product-to-sum identity
  • \( \sin A \sin B = \frac{1}{2}[\cos(A - B) - \cos(A + B)] \)
is particularly useful for converting the product of two sine functions into a sum, which is often easier to integrate. These identities are derived from the fundamental trigonometric properties and are essential tools in calculus, aiding in the handling and manipulation of trigonometric expressions.
product-to-sum
The product-to-sum identity is a powerful tool in simplifying integrals involving products of sine or cosine functions. This identity transforms a product into a sum or difference that can be more easily integrated. For example, in the exercise,
  • \( \int_{0}^{\pi} \sin 5x \sin 6x \, dx \)
is simplified using the identity,
  • \( \frac{1}{2}[\cos(x) - \cos(11x)] \),
turning a complex product into separate cosine functions. This transformation is advantageous because integrating cosine functions is more straightforward, as they have simple antiderivatives. Recognizing when to apply these identities can save significant time and effort during integration.
integration of trigonometric functions
Integrating trigonometric functions often involves applying specific techniques or identities to simplify the process. Functions such as sine and cosine have simple antiderivatives; however, when they appear in products or powers, more depth in manipulation is required. Consider
  • \( \int \cos(x) \, dx = \sin(x) + C \)
  • \( \int \cos(11x) \, dx = \frac{1}{11} \sin(11x) + C \)
The integral of a cosine function simply relates to the sine function, with adjustments for frequency if a coefficient is present. These straightforward steps are crucial in decomposing more complex integrals, as seen in the original exercise. Knowing these basic integrations is the foundation upon which more advanced calculus techniques are built.
integration limits
Definite integrals require evaluating the antiderivative at specific integration limits. This process determines the area under a curve within defined bounds. In the exercise, the limits are from 0 to \( \pi \). First, integrate the function, then evaluate the resulting expression at these limits:
  • \([\sin(x)]_{0}^{\pi} = \sin(\pi) - \sin(0) = 0\)
  • \([\sin(11x)]_{0}^{\pi} = \sin(11\pi) - \sin(0) = 0\)
These results show how trigonometric properties (like the periodicity of sine and cosine, where they return to zero after a full period like \( \pi \)) greatly simplify evaluations. Understanding and applying these properties is essential for efficiently calculating definite integrals, arriving at solutions that might initially seem complex but resolve neatly.

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Most popular questions from this chapter

The speed \(V(t) \mathrm{m} \mathrm{s}^{-1}\) of a vehicle at time \(t \mathrm{~s}\) is given by the table below. Use Simpson's rule to estimate the distance travelled over the eight seconds.$$ \begin{array}{l|lcccccccc} \hline t & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ V(t) & 0 & 0.63 & 2.52 & 5.41 & 9.02 & 13.11 & 16.72 & 18.75 & 20.15 \\ \hline \end{array} $$

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