Problem 41 - Calculus Single Variable

Question

Each of the integrands involves an expression of the form \(a^{2}-b^{2} x^{2}, a^{2}+b^{2} x^{2},\) or \(b^{2} x^{2}-a^{2} .\) Use an indirect substitution of the form \(x=(a / b) \sin (\theta), x=(a / b)\) \(\tan (\theta),\) or \(x=(a / b) \sec (\theta)\) to calculate the given integral. $$ \int_{0}^{1} \frac{3}{\sqrt{16+9 x^{2}}} d x $$

Step-by-Step Solution

Answer

The value of the integral is \( \ln(2) \).

Step 1: Recognize the integrand pattern

The integrand is of the form \( \frac{3}{\sqrt{a^2 + b^2 x^2}} \). Here, \( a^2 = 16 \) and \( b^2 = 9 \), leading to \( a = 4 \) and \( b = 3 \). This matches the \( a^2 + b^2 x^2 \) form.

Step 2: Choose the appropriate substitution

For integrals of the form \( a^2 + b^2 x^2 \), we use the substitution \( x = \frac{a}{b} \tan(\theta) \). Substituting, we have \( x = \frac{4}{3} \tan(\theta) \).

Step 3: Differentiate and find dx

Differentiate \( x = \frac{4}{3} \tan(\theta) \) to find \( dx \). This gives \( dx = \frac{4}{3} \sec^2 (\theta) d\theta \).

Step 4: Substitute into the integral

Substitute \( x = \frac{4}{3} \tan(\theta) \) and \( dx = \frac{4}{3} \sec^2(\theta) d\theta \) into the integral:\[ \int \frac{3}{\sqrt{16 + 9(\frac{4}{3} \tan(\theta))^2}} \times \frac{4}{3} \sec^2(\theta) d\theta \]

Step 5: Simplify the integrand

Simplify \( \sqrt{16 + 9(\frac{4}{3} \tan(\theta))^2} \):\( 16 + 9(\frac{16}{9} \tan^2(\theta)) = 16 + 16 \tan^2(\theta) = 16 \sec^2(\theta) \)So the integral becomes:\[ \int 3 \cdot \frac{4}{3} \cdot \frac{\sec^2(\theta)}{4 \sec(\theta)} d\theta \]

Step 6: Simplify and integrate

Further simplify:\[ \int \sec(\theta) d\theta \]which integrates to \( \ln |\sec(\theta) + \tan(\theta)| + C \).

Step 7: Back-substitute to original variable

Given that \( x = \frac{4}{3} \tan(\theta) \), then \( \tan(\theta) = \frac{3x}{4} \). Therefore, \( \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \frac{\sqrt{16 + 9x^2}}{4} \).The result is:\[ \ln \left| \frac{\sqrt{16 + 9x^2}}{4} + \frac{3x}{4} \right| + C \]

Step 8: Evaluate the definite integral

Use the limits \( x = 0 \) to \( x = 1 \):\[ \left[ \ln \left( \frac{\sqrt{16 + 9x^2} + 3x}{4} \right) \right]_0^1 \]Compute:At \( x = 1, \ln \left( \frac{\sqrt{25} + 3}{4} \right) = \ln(2) \)At \( x = 0, \ln(1) = 0 \)The integral evaluates to \( \ln(2) \).

Key Concepts

Substitution Method

When faced with complex integrals, the substitution method can transform them into simpler forms that are easier to solve. The idea is to replace a part of the integrand (the function inside the integral) with a new variable, which helps in tackling tricky expressions.

In this exercise, we specifically used a trigonometric form of substitution. Recognizing the pattern \( a^2 + b^2 x^2 \), we identified that a substitution involving tangent could simplify the expression. Trigonometric identities are often used in these cases since they can replace challenging algebraic forms, making integration manageable.

This method is useful not just for simplifying, but also for converting the integral into a common form that can be integrated directly or with minimal additional work.

Trigonometric Substitution

Trigonometric substitution is a specialized technique used in calculus to simplify integrals involving expressions like \( a^2 - b^2 x^2 \), \( a^2 + b^2 x^2 \), and \( b^2 x^2 - a^2 \). By substiting trigonometric functions for \( x \), the integration becomes more straightforward.

In our scenario, the expression matched the form \( a^2 + b^2 x^2 \), allowing us to apply \( x = \frac{a}{b} \tan(\theta) \). This effectively transforms the problem into a more standard form that many calculus students find easier to integrate.

Common substitutions include:

• \( x = \frac{a}{b} \sin(\theta) \) for \( a^2 - b^2 x^2 \)
• \( x = \frac{a}{b} \tan(\theta) \) for \( a^2 + b^2 x^2 \)
• \( x = \frac{a}{b} \sec(\theta) \) for \( b^2 x^2 - a^2 \)

Utilizing these substitutions, the process simplifies complex roots and enables the extraction of functions ready for integration.

Definite Integral Evaluation

Once substitution and simplification are complete, it's time to evaluate the definite integral. This involves taking the integral over a specified interval. In our exercise, the limits were from 0 to 1.

The goal is to transform the original variable limits into the new variable, effectively evaluating the antiderivative within the given bounds. It's crucial to substitute back into the original variable once the antiderivative is found and before applying the limits to ensure the evaluation pertains to the original function.

As seen, once integration is performed, we convert back to the variable \( x \), compute the antiderivative at both the upper and lower bounds, and find the difference. This absolute difference yields the precise numerical value of the integral, which, in our case, calculated to \( \ln(2) \).

These steps ensure that the solution not only solves the integral but does so exactly within its specified limits, an essential aspect of definite integrals.

Calculus Problem Solving

Calculus problem solving often involves a strategic approach, combining various techniques to reach the solution efficiently. The problem at hand included applying substitution and integration rules methodically to achieve the final result.

Problem solving begins with identifying patterns and structures within the integrand to select the appropriate method, like substitution or integration by parts. For this problem, identifying that trigonometric substitution simplifies the integral was key. Each step, from recognizing the pattern to back-substitution and evaluation, builds towards the final outcome.

Here are some strategies for effective calculus problem solving:

• Recognize standard forms and rules.
• Choose the simplest method applicable.
• Simplify expressions as much as possible.
• Carefully consider bounds in definite integrals.

Ultimately, combining logical reasoning with mathematical techniques is at the heart of solving complex calculus problems. Through practice, students can enhance their ability to navigate diverse calculus challenges efficiently.