Problem 16 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{2} \frac{s+1}{\sqrt{4-s^{2}}} d s $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{2} + 2 \).

Step 1: Identify the Integration Technique

The integral involves the term \( \frac{s+1}{\sqrt{4-s^{2}}} \). This suggests using a trigonometric substitution since \( 4 - s^2 \) is of the form \( a^2 - s^2 \). We can use the substitution \( s = 2\sin \theta \), which transforms \( \sqrt{4-s^2} \) into \( 2\cos \theta \).

Step 2: Substitute and Simplify

With the substitution \( s = 2\sin \theta \), then \( ds = 2\cos \theta \, d\theta \). The integral becomes:\[ \int \frac{(2\sin \theta + 1)}{2\cos \theta} \cdot 2\cos \theta \, d\theta \].This simplifies to \[ \int (2\sin \theta + 1) \, d\theta \].

Step 3: Integrate the Simplified Expression

Now integrate the simplified expression with respect to \( \theta \):\[ \int (2\sin \theta + 1) \, d\theta = 2\int \sin \theta \, d\theta + \int 1 \, d\theta \].The integral \( 2\int \sin \theta \, d\theta \) evaluates to \(-2\cos \theta \), and \( \int 1 \, d\theta \) evaluates to \( \theta \). Thus the integral becomes: \( -2\cos \theta + \theta \).

Step 4: Back-substitute to Original Variable

We now revert back to the variable \(s\) using the inverse substitutions. Since \( s = 2\sin \theta \), then \( \theta = \arcsin\frac{s}{2} \) and \( \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{s^2}{4}} = \frac{\sqrt{4-s^2}}{2} \). Therefore, \( \cos \theta = \frac{\sqrt{4-s^2}}{2} \) gives us the expression for the integral as: \( -2\left(\frac{\sqrt{4-s^2}}{2}\right) + \arcsin\frac{s}{2} \), which simplifies to \( -\sqrt{4-s^2} + \arcsin\frac{s}{2} \).

Step 5: Evaluate the Integral from 0 to 2

Evaluate \( -\sqrt{4-s^2} + \arcsin\frac{s}{2} \) at the bounds \( s = 0 \) to \( s = 2 \):- At \( s = 2 \), \(-\sqrt{4-2^2} + \arcsin\frac{2}{2} = -0 + \frac{\pi}{2} = \frac{\pi}{2} \).- At \( s = 0 \), \(-\sqrt{4-0^2} + \arcsin\frac{0}{2} = -2 + 0 = -2 \).The definite integral evaluates to \( \frac{\pi}{2} - (-2) = \frac{\pi}{2} + 2 \).

Key Concepts

Integral Calculus

Integral calculus is a crucial branch of mathematics focusing on the concept of integration. It revolves around finding the integral of functions, which essentially is the opposite process of differentiation. The main goal in integral calculus is to determine the total accumulation of quantities, such as areas under curves or total change over an interval.

When working with integrals, there are primarily two types to consider:

• Indefinite Integral: Represents a family of functions and includes a constant of integration.
• Definite Integral: Represents the exact area under a curve between two limits.

In our exercise, we utilized integral calculus to evaluate the expression involving the square root term. By breaking down the problem into manageable steps, integral calculus allows us to systematically find the area or evaluate the function over a specific interval.

Definite Integrals

Definite integrals are a type of integral used to compute the net area under a curve between two given points on the x-axis. In this context, we are given specific limits of integration, from 0 to 2 in the exercise. These limits define where the calculation starts and ends, and they help in expressing the exact value rather than a family of values.

The process involves substituting these limits into the antiderivative (the integral solution) to calculate the total area.
Here's the important part:

• First, evaluate the antiderivative expression at the upper limit.
• Subtract the evaluation of the antiderivative at the lower limit.

By following these steps, you can get an accurate numerical value that reflects the total change or area precisely. It’s also important to replace any trigonometric substitutions back to the original variable before plugging in these limits, as shown in the solution where we substituted back using inverse trigonometric functions.

Trigonometric Identities

Trigonometric identities often play a vital role in simplifying integrals that involve square roots or other complex expressions. In exercises similar to the one presented, trigonometric identities can transform complicated algebraic expressions into simpler forms, allowing for easier integration.

In our example, the substitution involved using identities like:

• The identity \( s = 2\sin \theta \) enabled us to change the form of \( \sqrt{4-s^2} \) into \( 2\cos \theta \, \text{since} \, \sin^2 \theta + \cos^2 \theta = 1 \).
• Another critical step was using \( \cos \theta = \sqrt{1- \sin^2 \theta} \), which is derived from the Pythagorean identity.

By employing these identities, the heavy lifting in integration was dramatically reduced, making the integration process much more straightforward. Understanding how to utilize these identities effectively is key to solving integrals involving trigonometric substitutions.