Problem 18 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int_{\pi / 2}^{\pi}(\sin y) e^{\cos y} d y $$

Step-by-Step Solution

Answer

The value of the integral is \( 1 - \frac{1}{e} \).

Step 1: Identify a Suitable Substitution

To simplify the integral \( \int_{\pi / 2}^{\pi} (\sin y) e^{\cos y} \, dy \), notice that the expression involves \( \sin y \) and \( e^{\cos y} \). A good substitution is \( u = \cos y \), because the derivative \( du = -\sin y \, dy \) is present in the integral, allowing us to substitute \( \sin y \, dy = -du \).

Step 2: Change the Limits of Integration

When \( y = \pi/2 \), \( \cos(\pi/2) = 0 \), so \( u = 0 \). When \( y = \pi \), \( \cos(\pi) = -1 \), so \( u = -1 \). Thus, the new limits are from \( u = 0 \) to \( u = -1 \).

Step 3: Substitute and Simplify the Integral

Substitute the expressions into the integral: \( \int_{\pi / 2}^{\pi} (\sin y) e^{\cos y} \, dy = \int_{0}^{-1} e^{u} (-du) \). Simplifying, this becomes \( -\int_{0}^{-1} e^{u} \, du \), which is equivalent to \( \int_{-1}^{0} e^{u} \, du \).

Step 4: Evaluate the Integral

The integral of \( e^{u} \) is itself. Thus, evaluate \( \int_{-1}^{0} e^{u} \, du \) as \( [e^{u}]_{-1}^{0} = e^{0} - e^{-1} = 1 - \frac{1}{e} \).

Step 5: Write the Final Answer

After evaluating, the final result of the integral is \( 1 - \frac{1}{e} \).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique in calculus. It simplifies integrals by changing variables, making the integral easier to solve. This method is especially useful when dealing with complex expressions. In this example, we use substitution to solve the integral \( \int_{\pi / 2}^{\pi}(\sin y) e^{\cos y} \, d y \).

The key is to find a substitution that will convert the integral into a simpler form. We start by examining the integral and identify that \( u = \cos y \) is a suitable substitution. The derivative \( du = -\sin y \, dy \) closely matches the integral's elements, allowing us to make the transformation.

This technique changes the problem's limits and the integrand into an easier function to evaluate. Ultimately, substitution transforms a daunting integral into a standard one, easing the integration process.

Definite Integrals

Definite integrals are used to calculate the area under a curve between two limits. These integrals have both an upper and lower bound. For the given problem, the upper and lower bounds are \( y = \pi \) and \( y = \pi/2 \), respectively.

After applying integration by substitution, these limits must be converted to match the new variable. Substituting \( u = \cos y \) changes the limits as follows:

• When \( y = \pi/2 \), \( u = \cos(\pi/2) = 0 \)
• When \( y = \pi \), \( u = \cos(\pi) = -1 \)

Consequently, the limits for \( u \) become \( u = 0 \) to \( u = -1 \).

Definite integrals provide a precise numerical value as opposed to an indefinite integral, which results in a general formula. Therefore, as seen in this example, the simplified integral can be evaluated directly to find the area.

Exponential Functions

Exponential functions play a crucial role in calculus due to their unique properties, particularly in integrals. The integral \( \int_{0}^{-1} e^{u} \, du \) demonstrates a typical exponential function scenario.

In these functions, the rate of increase is proportional to the value of the function itself. The exponential function is defined as \( e^{x} \), where \( e \) is Euler's number, approximately 2.718. Notably, the derivative and integral of \( e^{x} \) are both \( e^{x} \), highlighting their simplicity in calculus operations.

This property makes them very predictable and easier to handle in integrals. In our example, integrating \( e^{u} \) over the interval \( -1 \) to \( 0 \) is straightforward. The result is \( e^0 - e^{-1} = 1 - \frac{1}{e} \), showcasing how exponential functions can simplify integration solutions.

Trigonometric Functions

Trigonometric functions such as sine and cosine are essential in calculus due to their periodic nature and occurrence in different problems. In the given integral, \( \int_{\pi / 2}^{\pi}(\sin y) e^{\cos y} \, d y \), the trigonometric function \( \cos y \) helps establish the substitution.

Trigonometric identities and derivatives are utilized to transform the integral for easier calculation. Here:

• \( \sin y \) acts as a part of \( du \) in the substitution process, given \( du = -\sin y \, dy \)
• \( \cos y \) is used as the substitution variable \( u \)

Trigonometric functions’ derivatives offer flexibility for solving complex integrals, as they provide a bridge between seemingly unrelated terms.

This example shows how powerful trigonometric functions can be when used within substitution and integration hyperfunctions, breaking down multifaceted functions into simpler expressions to evaluate.