Chapter 6: Problem 12
\(\int_{0}^{\pi / 4} \sin 2 \theta d \theta=\) (A) 2 (B) \(\frac{1}{2}\) (C) -1 (D) \(-\frac{1}{2}\)
Short Answer
Expert verified
The integral evaluates to \(-\frac{1}{2}\), which is option (D).
Step by step solution
01
Identify the Integrand and Limits
The problem requires evaluating the definite integral \( \int_{0}^{\pi / 4} \sin 2\theta \, d\theta \). Here, the integrand is \( \sin 2\theta \) and the limits of integration are from \( 0 \) to \( \pi/4 \).
02
Use a Substitution for Simplification
To make this integration easier, use the substitution \( u = 2\theta \). Then, \( du = 2 \, d\theta \) or \( d\theta = \frac{1}{2} \, du \). Also, transform the limits: when \( \theta = 0 \), \( u = 0 \); and when \( \theta = \pi/4 \), \( u = \pi/2 \).
03
Rewrite the Integral with Substitution
With the substitution in place, rewrite the integral: \( \int_{0}^{\pi/4} \sin 2\theta \, d\theta = \int_{0}^{\pi/2} \sin u \cdot \frac{1}{2} \, du \). This simplifies to \( \frac{1}{2} \int_{0}^{\pi/2} \sin u \, du \).
04
Integrate the Transformed Integral
Find the antiderivative of \( \sin u \), which is \(-\cos u \). Applying the limits gives: \( \frac{1}{2} \left[ -\cos u \right]_{0}^{\pi/2} \).
05
Evaluate the Definite Integral
Substitute the limits into the antiderivative expression: \( \frac{1}{2} \left( -\cos(\pi/2) + \cos(0) \right) \). Since \( \cos(\pi/2) = 0 \) and \( \cos(0) = 1 \), we have \( \frac{1}{2} \left( 0 - 1 \right) = -\frac{1}{2} \).
06
Match Result to the Given Options
The computed result of the integral is \(-\frac{1}{2}\), which corresponds to option (D) in the given choices.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
In calculus, a definite integral is a fundamental concept involving the accumulation of quantities over a specific interval.
This is represented as the integral of a function between two points on its domain, known as the lower and upper limits. For example, in the exercise above, the definite integral is written as
\( \int_{0}^{\pi / 4} \sin 2\theta \ d \theta \). The function \( \sin 2\theta \) is our integrand, and the limits are from 0 to \( \pi/4 \).
This is represented as the integral of a function between two points on its domain, known as the lower and upper limits. For example, in the exercise above, the definite integral is written as
\( \int_{0}^{\pi / 4} \sin 2\theta \ d \theta \). The function \( \sin 2\theta \) is our integrand, and the limits are from 0 to \( \pi/4 \).
- The lower limit, 0, and the upper limit, \( \pi/4 \), define the interval along the x-axis over which the function is to be integrated.
- The integral sums up the area under the curve \( \sin 2\theta \) from 0 to \( \pi/4 \).
Integration Techniques
Integration is the process of finding the antiderivative of a function, and various techniques can be employed to achieve this. These methods simplify functions to make integration more straightforward:
Different functions require different strategies for integration, and using the right one is crucial.
Different functions require different strategies for integration, and using the right one is crucial.
- Substitution Method: In the problem given, substitution is used to simplify the integration since the integrand is \( \sin 2\theta \). By setting \( u = 2\theta \), it makes the integration more manageable. After substituting, the integral becomes \( \frac{1}{2} \int \sin u \ du \).
- Recognizing Patterns: Identifying basic derivatives and algebraic manipulations are always a good start. For example, knowing the integral of \( \sin u \) is \( -\cos u \).
Trigonometric Substitution
Trigonometric substitution is a powerful technique used when the integrand involves roots or trigonometric expressions.
It helps in transforming complex expressions into simpler trigonometric forms:
It helps in transforming complex expressions into simpler trigonometric forms:
- This was particularly useful in situations involving identities and transformations, such as in our exercise where \( \sin 2\theta \) was involved.
- Sometimes, if the integrand displays a trigonometric function, a substitution can exploit trigonometric identities to ease the process. In this case, a simple substitution \( u = 2\theta \) turned the integral into a form involving \( \sin u \), which is straightforward to solve.