Problem 13 - Essential Calculus

Question

Use the Special Integration Formulas (Theorem 6.2) to find the integral. $$ \int \sqrt{25-4 x^{2}} d x $$

Step-by-Step Solution

Answer

The solution to the integral is \(x \sqrt{25 - 4x^2} + 12.5 \sin^{-1} \left(\frac{2x}{5}\right) + C\)

Step 1: Identify the type of the integral

First, we need to recognize the integral \[\int \sqrt{25-4 x^{2}} dx\] as being in the form of \(\int \sqrt{a^2 - x^2} dx\). In this case, we have \(a = 5\) and \(x^2 = (2x)^2\). Therefore, our integral can be rewritten as \[\int \sqrt{25-(2x)^{2}} dx\]

Step 2: Apply the formula

Now we apply the referenced special integration formula. This will form a new equation: \[\frac{x}{2} \sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \left(\frac{x}{a}\right) + C\] Plugging our numbers into the formula will result in: \[\frac{(2x)}{2} \sqrt{5^2 - (2x)^2} + \frac{5^2}{2} \sin^{-1} \left(\frac{2x}{5}\right) + C\] which simplifies to: \[x \sqrt{25 - 4x^2} + 12.5 \sin^{-1} \left(\frac{2x}{5}\right) + C\]

Step 3: Simplify the formula

It is vital to ensure that everything is simplified to the best extent it can be. Hence, the solution to the integral \[\int \sqrt{25-4 x^{2}} d x\] is \[x \sqrt{25 - 4x^2} + 12.5 \sin^{-1} \left(\frac{2x}{5}\right) + C\]

Key Concepts

Definite and Indefinite Integrals

Integrals are fundamental tools in calculus that help us calculate the area under a curve.
Indefinite integrals, also known as antiderivatives, represent a family of functions and include a constant of integration, denoted as "+ C." This reflects the idea that when you take an antiderivative, you can add any constant. The integral of a function generally takes the form \( \int f(x) \, dx = F(x) + C \), where \( F'(x) = f(x) \). This notation indicates an indefinite integral since it does not have specified limits.

Definite integrals, on the other hand, calculate the actual area under a curve between specified limits \( a \) and \( b \). In definite integrals, the notation is \( \int_{a}^{b} f(x) \, dx \). Calculating the definite integral gives you a numerical value, which corresponds to the total area under the curve from \( x = a \) to \( x = b \). Unlike indefinite integrals, definite integrals do not include a "+ C" term.

Recognizing when to apply indefinite or definite integrals is crucial in solving calculus problems effectively. It hinges on whether you are asked to find a general expression or a specific numerical value of area.

Trigonometric Substitution

Trigonometric substitution is a clever technique used in calculus to simplify integrals involving root expressions such as \( \sqrt{a^2 - x^2} \). This method uses trigonometric identities to transform the integral into a more manageable form.
For the integral \( \int \sqrt{25 - 4x^2} \, dx \), we made use of the substitution \( x = \frac{5}{2} \sin \theta \), which transforms the expression under the square root into a trigonometric function.

Here's the key idea:

• For \( \sqrt{a^2 - x^2} \), substitute \( x = a \sin \theta \)
• For \( \sqrt{a^2 + x^2} \), substitute \( x = a \tan \theta \)
• For \( \sqrt{x^2 - a^2} \), substitute \( x = a \sec \theta \)

This substitution allows the powers and radicals to simplify, making the integral solvable using these trigonometric properties. After substitution, it’s essential to remember to revert back to the original variable \( x \) by using the inverse trigonometric function.

Inverse Trigonometric Functions

Inverse trigonometric functions play a pivotal role in reversing trigonometric relationships. In calculus, they are particularly useful for integrating expressions where standard trigonometric substitutions have been applied. For example, if you used \( x = 5 \sin \theta \), you can solve for \( \theta \) by finding \( \theta = \sin^{-1} \left( \frac{x}{5} \right) \).
These functions provide exact values for angles when given coordinate ratios. Key inverse trigonometric functions include:

• \( \sin^{-1}(x) \) - Gives the angle whose sine is \( x \)
• \( \cos^{-1}(x) \) - Provides the angle whose cosine is \( x \)
• \( \tan^{-1}(x) \) - Finds the angle whose tangent is \( x \)

The role of inverse trigonometric functions becomes especially clear in integration problems where they help express the resulting angles in terms of the original variable \( x \). In the solution, you’ll notice \( \sin^{-1} \left( \frac{2x}{5} \right) \) appearing as part of the product of integration, highlighting how trigonometric transformations are reversed.