Problem 16 - Calculus for Biology and Medicine

Question

Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int \frac{1}{\sqrt{16-9 x^{2}}} d x $$

Step-by-Step Solution

Answer

\( \int \frac{1}{\sqrt{16-9x^2}} \, dx = \frac{1}{3} \arcsin\left(\frac{3x}{4}\right) + C \)

Step 1: Identify the Integral Form

The integral given is \( \int \frac{1}{\sqrt{16-9x^2}} \, dx \). This integral resembles a form that can be simplified using trigonometric substitution, specifically, it looks like \( \int \frac{1}{\sqrt{a^2 - (bx)^2}} \, dx \). In this case, \( a = 4 \) and \( b = 3 \). The form matches \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \).

Step 2: Use a Suitable Substitution

Since it matches \( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx \), apply the substitution \( x = \frac{a}{b} \sin \theta = \frac{4}{3} \sin \theta \). This gives us \( dx = \frac{4}{3} \cos \theta \, d\theta \). Substitute into the integral to transform it into terms of \( \theta \).

Step 3: Transform and Simplify the Integral

Substitute \( x = \frac{4}{3} \sin \theta \) into the integral: \[ \int \frac{1}{\sqrt{16 - 9(\frac{4}{3} \sin \theta)^2}} \cdot \frac{4}{3} \cos \theta \, d\theta \] Simplify inside the square root: \[ \int \frac{4\cos \theta}{3\sqrt{16 - 16\sin^2 \theta}} \, d\theta \] Use \( 1 - \sin^2 \theta = \cos^2 \theta \), therefore the expression becomes:\[ \int \cos \theta \, d\theta \]

Step 4: Integrate and Back-Substitute

Integrate \( \int \cos \theta \, d\theta = \sin \theta + C \). Now substitute back \( \theta \) terms by using \( \sin \theta = \frac{3}{4}x \), thus \( \theta = \arcsin(\frac{3}{4}x) \). Therefore, the integral evaluates to \( \sin(\arcsin(\frac{3}{4}x)) = \frac{3}{4}x \).

Step 5: Final Expression of Integration

The final expression with back-substitution gives:\[ \int \frac{1}{\sqrt{16-9x^2}} \, dx = \frac{1}{3} \arcsin\left(\frac{3x}{4}\right) + C \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify integrals involving square roots. This method works particularly well when the integrand contains expressions in the form \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2}\). By using a trigonometric identity, we can transform these square roots into simpler trigonometric functions.

This substitution exploits the relationships between trigonometric functions and right triangles. For example, when you have \(\sqrt{a^2 - x^2}\), you might use \(x = a \sin \theta\). The reason this works is because the identity \(1 - \sin^2 \theta = \cos^2 \theta\) allows us to eliminate the square root and deal with a simpler form, such as \cos(\theta)\.

• Choose the correct substitution: \(x = a \sin \theta\), \(x = a \tan \theta\), or \(x = a \sec \theta\).
• Compute \(dx\) using the derivative of the substitution.
• Replace all instances of \(x\) and \(dx\) with expressions in \(\theta\).
• Integrate with respect to \(\theta\), then back-substitute to find the answer in terms of \(x\).

Definite Integrals

A definite integral is a type of integral that calculates the net area under a curve between two specific points. It provides a numerical value that represents the accumulation of quantities, such as length, area, or volume, over an interval \[a, b\]. In notation, it is represented as \int_a^b f(x) \, dx\.

One of the main properties of definite integrals is their dependency on limits of integration. The result of a definite integral is affected by the specific values of \(a\) and \(b\), unlike indefinite integrals, which provide a family of functions without a specific solution. When finding the definite integral, you apply the following steps:

• Find the indefinite integral of the function.
• Evaluate the antiderivative at the upper limit and subtract the value of the antiderivative evaluated at the lower limit using \[F(b) - F(a)\].
• Interpret the result in context, ensuring it aligns with the problem at hand.

This process reveals the **net area** between the function graph and the x-axis, considering above-axis values as positive and below-axis values as negative contributions.

Indefinite Integrals

Indefinite integrals represent a family of functions and are the opposite of taking the derivative. Indefinite integration of a function \(f(x)\) produces an antiderivative plus a constant \(C\). The constant is essential because any number of constants can create a family of curves that are vertically shifted versions of one another.

The general notation for an indefinite integral is \int f(x) \, dx = F(x) + C\, where \(F(x)\) is an antiderivative of \(f(x)\). To solve indefinite integrals, follow these steps:

• Identify the function \(f(x)\) whose integral you need to determine.
• Use known integration techniques, such as direct antiderivatives, substitution, or integration by parts, to find \(F(x)\).
• Add the constant \(C\) to the solution.

The concept of indefinite integrals is foundational in calculus and is crucial for solving differential equations and understanding continuous accumulation processes. Unlike definite integrals, they lack specific bounds and therefore represent a **general solution** applicable to a broad range of values.