Problem 52
Question
Additional integrals Evaluate the following integrals. $$\int_{\pi / 6}^{\pi / 2} \frac{d y}{\sin y}$$
Step-by-Step Solution
Verified Answer
The value of the integral is $$-\log |2 - \frac{\sqrt{3}}{3}|$$.
1Step 1: Identify the antiderivative
In this step, we identify the antiderivative of $$\frac{1}{\sin y}$$ which is $$\log |\csc y - \cot y|$$.
2Step 2: Apply the limits of integration
Now, we can use the Fundamental Theorem of Calculus to evaluate the integral $$\int_{\pi / 6}^{\pi / 2} \frac{d y}{\sin y}$$.
Using the antiderivative, we have:
$$\int_{\pi / 6}^{\pi / 2} \frac{d y}{\sin y} = \log |\csc y - \cot y| \Big|_{\pi/6}^{\pi/2}$$
3Step 3: Evaluate the antiderivative at the upper limit of integration
Put y = π/2 in $$\log |\csc y - \cot y|$$ and find the value:
$$\log |\csc(\pi/2) - \cot(\pi/2)| = \log | 1 - 0| = \log 1$$
4Step 4: Evaluate the antiderivative at the lower limit of integration
Put y = π/6 in $$\log|\csc y - \cot y|$$ and find the value:
$$\log |\csc(\pi/6) - \cot(\pi/6)| = \log |2 - \frac{\sqrt{3}}{3}|$$
5Step 5: Subtract the results of Step 3 and Step 4 to find the integral
Subtract the result from Step 4 from the result of Step 3 to find the value of the integral:
$$\int_{\pi / 6}^{\pi / 2} \frac{d y}{\sin y} = \log 1 - \log |2 - \frac{\sqrt{3}}{3}| = -\log |2 - \frac{\sqrt{3}}{3}|$$
So, the value of the integral is $$-\log |2 - \frac{\sqrt{3}}{3}|$$.
Key Concepts
AntiderivativeFundamental Theorem of CalculusLimits of IntegrationTrigonometric Integrals
Antiderivative
Understanding the concept of an antiderivative is crucial in calculus. It’s simply the reverse process of differentiation. If taking a derivative of a function gives us the rate at which something changes, finding an antiderivative allows us to reconstruct the original function from its rate of change.
For example, if you know that the rate at which a car's position is changing (its speed) is constant, then by finding an antiderivative, you can determine the car's position at any point in time. In calculus, we often represent the antiderivative with an integral symbol and use it to find the area under a curve or solve problems involving motion.
In the given integral, the antiderivative of \(\frac{1}{\text{sin} y}\) is \(\text{log} |\text{csc} y - \text{cot} y|\), which represents the function whose rate of change is \(\frac{1}{\text{sin} y}\). Identifying the correct antiderivative is the first step towards solving an integral.
For example, if you know that the rate at which a car's position is changing (its speed) is constant, then by finding an antiderivative, you can determine the car's position at any point in time. In calculus, we often represent the antiderivative with an integral symbol and use it to find the area under a curve or solve problems involving motion.
In the given integral, the antiderivative of \(\frac{1}{\text{sin} y}\) is \(\text{log} |\text{csc} y - \text{cot} y|\), which represents the function whose rate of change is \(\frac{1}{\text{sin} y}\). Identifying the correct antiderivative is the first step towards solving an integral.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is one of the most important principles in mathematics, bridging the gap between differentiation and integration. It establishes the relationship between the derivative and the integral of a function and consists of two parts:
The first part guarantees that the integral of a function over a certain interval can be computed using any of its infinitely many antiderivatives. This part is essentially what allows us to evaluate definite integrals.
The second part states that the derivative of the integral of a function is the original function itself. It ensures that differentiation and integration are inverse processes. For solving the integral in question, we use the first part of the theorem to evaluate the integral at the bounds defined by the limits of integration.
The first part guarantees that the integral of a function over a certain interval can be computed using any of its infinitely many antiderivatives. This part is essentially what allows us to evaluate definite integrals.
The second part states that the derivative of the integral of a function is the original function itself. It ensures that differentiation and integration are inverse processes. For solving the integral in question, we use the first part of the theorem to evaluate the integral at the bounds defined by the limits of integration.
Limits of Integration
Limits of integration specify the interval over which we are integrating a function. In a definite integral, such as \(\int_{a}^{b} f(x) \, dx\), the lower limit of integration is 'a' and the upper limit is 'b'. These limits describe the boundaries of the area under the curve of the function f(x) within a specific range on the x-axis.
Having correct limits of integration is essential for accurately determining the value of a definite integral. In the provided exercise, the limits of integration are \(\frac{\pi}{6}\) and \(\frac{\pi}{2}\), which tells us that we're looking at the behavior of our function (\(\frac{1}{\sin y}\)) when y is between these two points of the trigonometric circle.
Having correct limits of integration is essential for accurately determining the value of a definite integral. In the provided exercise, the limits of integration are \(\frac{\pi}{6}\) and \(\frac{\pi}{2}\), which tells us that we're looking at the behavior of our function (\(\frac{1}{\sin y}\)) when y is between these two points of the trigonometric circle.
Trigonometric Integrals
Trigonometric integrals involve integrating functions with trigonometric expressions. They often require the use of trigonometric identities to simplify the integrand before integrating. For instance, integrating \(\frac{1}{\sin y}\) requires recognizing that this is the csc(y), which is not straightforward to integrate.
To solve these types of integrals, we sometimes express them in terms of other trigonometric functions to find an antiderivative that is easier to work with. This often involves strategies such as trigonometric substitutions or using identities such as \(1 + \cot^2 y = \csc^2 y\). In our exercise, we used the antiderivative \(\log |\csc y - \cot y|\) to evaluate the integral from \(\frac{\pi}{6}\) to \(\frac{\pi}{2}\), which is a common technique for integrals involving csc(y) or sec(y).
To solve these types of integrals, we sometimes express them in terms of other trigonometric functions to find an antiderivative that is easier to work with. This often involves strategies such as trigonometric substitutions or using identities such as \(1 + \cot^2 y = \csc^2 y\). In our exercise, we used the antiderivative \(\log |\csc y - \cot y|\) to evaluate the integral from \(\frac{\pi}{6}\) to \(\frac{\pi}{2}\), which is a common technique for integrals involving csc(y) or sec(y).
Other exercises in this chapter
Problem 52
Use a computer algebra system to evaluate the following indefinite integrals. Assume that a is a positive real number. $$\int \frac{d x}{x\left(a^{2}-x^{2}\righ
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Evaluate the following definite $$\int_{\sqrt{2}}^{2} \frac{\sqrt{x^{2}-1}}{x} d x$$
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Use a substitution to reduce the following integrals to \(\int\) ln \(u\) du. Then evaluate the resulting integral. $$\int(\cos x) \ln (\sin x) d x$$
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Use the approaches discussed in this section to evaluate the following integrals. $$\int_{0}^{\pi / 8} \sqrt{1-\cos 4 x} d x$$
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