Problem 47 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(29-50.\) $$\int_{\ln (\pi / 6)}^{\ln (\pi / 2)} 2 e^{v} \cos e^{v} d v$$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: Recognize the Substitution

We notice that the integral has the form \( \int 2 e^v \cos(e^v) \, dv \). This suggests a substitution method might work well. By closely looking at the expression, we can use the substitution \( u = e^v \), which implies \( du = e^v \, dv \).

Step 2: Adjust the Limits of Integration

With the substitution \( u = e^v \), we need to adjust the integration limits. When \( v = \ln(\pi/6) \), then \( u = e^{\ln(\pi/6)} = \pi/6 \). Similarly, when \( v = \ln(\pi/2) \), then \( u = e^{\ln(\pi/2)} = \pi/2 \).

Step 3: Replace Variables and Simplify the Integral

The integral \( \int 2 e^v \cos(e^v) \, dv \) becomes \( \int_{\pi/6}^{\pi/2} 2 \cos(u) \, du \) after the substitution \( u = e^v \) and setting the new limits.

Step 4: Integrate with Respect to New Variable

We now integrate \( \int 2 \cos(u) \, du \). The integral of \( \cos(u) \) is \( \sin(u) \), so we have \[ 2 \int \cos(u) \, du = 2 \sin(u) \].

Step 5: Evaluate the Definite Integral

Now, with the expression \( 2 \sin(u) \), we evaluate from \( u = \pi/6 \) to \( u = \pi/2 \). This gives us \[ 2 \sin(\pi/2) - 2 \sin(\pi/6) \].

Step 6: Calculate Values of Trigonometric Functions

Calculate the sine values: \( \sin(\pi/2) = 1 \) and \( \sin(\pi/6) = 1/2 \). So, \( 2 \sin(\pi/2) - 2 \sin(\pi/6) = 2(1) - 2(1/2) = 2 - 1 = 1 \).

Key Concepts

Definite Integrals

In calculus, a definite integral is a way to find the area under a curve between two specific points, known as the limits of integration. Unlike indefinite integrals, which provide a general formula, definite integrals result in a number.
The notation for a definite integral is \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits, respectively. Evaluating it involves finding the antiderivative of the function \( f(x) \), substituting the limits into this antiderivative, and finding the difference between the two.
This process transforms complex curves into tangible areas, allowing for more practical applications.

Substitution Method

The substitution method is a powerful tool in solving integrals, especially when dealing with composite functions. By using an appropriate substitution, you can simplify the integral to a more basic form, making it easier to evaluate.

• First, identify a part of the integral you can replace with a single variable (such as \( u \)).
• Then, express the entire integral in terms of this new variable.
• Don't forget to modify the differential \( dv \) accordingly, by computing \( du \).
• Update the limits of integration to reflect the new variable \( u \).

The substitution \( u = e^v \) in the original problem allows us to transform a complex integral involving \( e^v \) into a simpler integral in terms of \( u \), making it more manageable to solve.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are fundamental in integral calculus. They are periodic functions that model wave-like behaviors and are essential in various applications.
When integrating trigonometric functions, it is helpful to remember their basic antiderivatives:

• The antiderivative of \( \sin(x) \) is \( -\cos(x) \).
• The antiderivative of \( \cos(x) \) is \( \sin(x) \).

In the context of definite integrals, as seen in the problem, the integration of \( \cos(u) \) over a certain range is straightforward, yielding \( \sin(u) \). Evaluating this result within specified limits provides a precise area or value, such as the simple arithmetic leading to the final solution.