Problem 56 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{0}^{\pi / 2} 2 \sinh (\sin \theta) \cos \theta d \theta$$

Step-by-Step Solution

Answer

The value of the integral is \( e + e^{-1} - 2 \).

Step 1: Identify the Integral Type

We are asked to evaluate the integral \( \int_{0}^{\pi / 2} 2 \sinh (\sin \theta) \cos \theta \, d\theta \). Noticing the composition of \( \sinh \) and \( \sin \) suggests that a substitution might simplify the integration.

Step 2: Apply Substitution

Let's use the substitution \( u = \sin \theta \). Then we have \( du = \cos \theta \, d\theta \). The limits of integration change accordingly: when \( \theta = 0 \), \( u = \sin(0) = 0 \) and when \( \theta = \frac{\pi}{2} \), \( u = \sin\left(\frac{\pi}{2}\right) = 1 \). The integral becomes \( \int_{0}^{1} 2 \sinh(u) \, du \).

Step 3: Integrate with Respect to u

To evaluate the new integral, use the fact that the integral of \( \sinh(u) \) is \( \cosh(u) \). Thus, the integral \( \int 2 \sinh(u) \, du = 2 \cosh(u) + C \).

Step 4: Apply the Fundamental Theorem of Calculus

Now we apply the limits of integration to the antiderivative. Evaluate \( 2 \cosh(u) \) from 0 to 1, which gives \( 2 \cosh(1) - 2 \cosh(0) \).

Step 5: Calculate the Value

We know that \( \cosh(0) = 1 \) and \( \cosh(1) = \frac{e + e^{-1}}{2} \). Thus, the result is \( 2 \left( \frac{e + e^{-1}}{2} \right) - 2 \times 1 = e + e^{-1} - 2 \).

Key Concepts

Definite Integrals

A definite integral is a beautiful concept in calculus that helps calculate the area under the curve of a function between two specific points. It's not just about finding an area; it's more like solving a puzzle where you fit tiny rectangles under a curve to see the bigger picture. Here's the quick rundown of how definite integrals work:

• They involve integration over a specific interval, say from \( a \) to \( b \), denoted as \( \int_{a}^{b} f(x) \, dx \).
• The result is a number, not a function.
• You apply the Fundamental Theorem of Calculus, which connects differentiation and integration -- they are inverses of each other!

In our exercise, we found the definite integral of a function that involves hyperbolic and trigonometric functions. We went from \( \theta = 0 \) to \( \theta = \pi/2 \) and transformed this into another variable through substitution. This allowed us to apply the theorem efficiently and find the net area or accumulation between those points.

Hyperbolic Functions

Hyperbolic functions often resemble trigonometric functions but are deeply rooted in exponential behaviors. Just like trigonometric functions relate to circles, hyperbolic functions relate to hyperbolas. Let's peek into the world of hyperbolic functions with some key points:

• Two primary hyperbolic functions are \( \sinh(x) \) and \( \cosh(x) \).
• The definition revolves around the exponential function: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \) and \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).
• Interestingly, these functions help describe real-life phenomena, such as hanging cables, or the shape of a satellite dish.

In our case, we evaluated the integral of \( 2 \sinh(u) \), a hyperbolic function. By integrating \( \sinh(u) \), we found the result is \( 2 \cosh(u) \). This transformation is essential for finding the definite integral, which leads to the final answer. Understand these functions well, and you'll see how they elegantly simplify many complex integrals.

Trigonometric Substitution

The trick of trigonometric substitution simplifies integrals that involve expressions similar to trigonometric identities. Imagine having a complex expression transformed into something familiar and solvable through a smart substitution! Here's how trigonometric substitution aids in our journey:

• It's useful when you see the product of trigonometric functions or even several algebraic expressions inside an integral.
• The goal is to switch to a trigonometric variable so integral evaluations become easier.
• Once substituted, you translate the limits of integration to fit the new variable.

For our exercise, we substituted \( u = \sin \theta \), which turned the integration problem into evaluating \( \int_{0}^{1} 2 \sinh(u) \, du \). This use of trigonometric substitution simplified the complex form and converted it into a doable integral of hyperbolic functions. Mastering this strategy opens doors to tackle a variety of challenging integration problems.