Problem 54 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int_{-2 / 3}^{-\sqrt{2} / 3} \frac{d y}{y \sqrt{9 y^{2}-1}}$$

Step-by-Step Solution

Answer

\( \frac{\sqrt{2} - \sqrt{3}}{2} \)

Step 1: Substitution Selection

To evaluate the integral \( \int \frac{dy}{y \sqrt{9y^2 - 1}} \), we select the substitution that will simplify the square root term. Notice that this integral has the form similar to \( \int \frac{dy}{y \sqrt{ay^2 - b}} \). The substitution \( y = \frac{1}{3} \sec \theta \) can simplify the expression, where \( ay^2 - b \to \sec^2 \theta - 1 = \tan^2 \theta \).

Step 2: Applying Substitution

Substitute \( y = \frac{1}{3} \sec \theta \), then \( dy = \frac{1}{3} \sec \theta \tan \theta \, d\theta \). The limits of \( y \) will change to limits of \( \theta \). When \( y = -2/3, \sec \theta = -2 \), gives \( \theta = \cos^{-1}(-1/2) = 2\pi/3 \). Similarly, when \( y = -\sqrt{2}/3, \sec \theta = -\sqrt{2} \), gives \( \theta = \cos^{-1}(-\sqrt{1/2}) = 3\pi/4 \).

Step 3: Simplifying the Integral

The integral becomes \( \int_{2\pi/3}^{3\pi/4} \frac{1}{\frac{1}{3} \sec \theta \sqrt{9 (\frac{1}{3} \sec \theta)^2 - 1}} \cdot \frac{1}{3} \sec \theta \tan \theta \, d\theta \). This simplifies to \( \int_{2\pi/3}^{3\pi/4} \frac{1}{\tan \theta} d\theta = \int_{2\pi/3}^{3\pi/4} \cos \theta d\theta \).

Step 4: Evaluate the Simplified Integral

The integral \( \int \cos \theta \, d\theta \) is solved as \( \sin \theta \). Evaluating from \( \theta = 2\pi/3 \) to \( \theta = 3\pi/4 \), \[ \sin(3\pi/4) - \sin(2\pi/3). \] Calculate these values: \( \sin 3\pi/4 = \sin \pi/4 = \frac{\sqrt{2}}{2} \) and \( \sin 2\pi/3 = \sin \pi/3 = \frac{\sqrt{3}}{2} \). The result is \[ \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}. \]

Step 5: Provide the Final Answer

After evaluating, the integral \( \int_{-2 / 3}^{-\sqrt{2} / 3} \frac{d y}{y \sqrt{9 y^{2}-1}} \) evaluates to \( \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}. \) Thus, the final answer is \[ \frac{\sqrt{2} - \sqrt{3}}{2}. \]

Key Concepts

Definite Integrals

Definite integrals represent the accumulation of quantities over an interval. Unlike indefinite integrals, which give a family of functions, definite integrals result in a specific numerical value. This value often represents area under a curve.

In the exercise, the definite integral is evaluated from \(-2/3\) to \(-\sqrt{2}/3\). This means we are calculating the accumulation of the function \(\frac{1}{y \sqrt{9y^2 - 1}}\) over this specific interval. The result is the difference between the values of the antiderivative at the upper limit and lower limit.

This process involves several steps:

• Selecting appropriate substitution to simplify the integral.
• Transforming the limits of integration according to the substitution.
• Performing the integration.
• Substituting back to original variables (if necessary) before evaluating.

Incorporating substitution changes the limits from \(y\)-values to corresponding \(\theta\)-values, keeping the integral in terms of the trigonometric function aside from achieving simplicity in calculations.

Trigonometric Identities

Trigonometric identities are mathematical equations involving trigonometric functions. These identities help simplify complex trigonometric expressions, which is vital in calculus, particularly in integration.

Common identities include:

• The Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Secant and tangent identity: \(\sec^2 \theta = 1 + \tan^2 \theta\)
• Angle sum formulae

In the provided solution, the substitution \(y = \frac{1}{3} \sec \theta\) transforms the integral to terms of \(\theta\). Through \(\sec^2 \theta - 1 = \tan^2 \theta\), the square root expression simplifies, and using \(\tan\) and \(\sec\) identities helps us further reduce the complexity of the integral.

This elimination of radicals allows the integrand to be presented in terms of simpler trigonometric functions, eventually leading to integration and back substitution seamlessly.

Calculus Integration Techniques

Calculus integration techniques are strategies or methods used to find the antiderivative or area under the curve for functions that are quite complex in their algebraic form.

Techniques include:

• Substitution method: Simplifying the integral by substituting parts of the integrand with new variables.
• Integration by parts: Used often when integrands are products of functions.
• Partial fraction decomposition: Breaking down rational functions into simpler fractions.

The exercise leverages the substitution method extensively. By selecting \(y = \frac{1}{3} \sec \theta\), the initially complex integral expression simplifies after transforming it to \(\theta\)-terms. Consequently, integration becomes manageable, and you execute the process by evaluating simpler trigonometric functions.

Ultimately, these techniques are designed to transform intimidating expressions into forms that are easier to integrate, showcasing the power and utility of calculus when dealing with difficult mathematical problems.