Problem 39 - 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. $$ \int_{0}^{1 / \sqrt{2}} \frac{1}{16 x^{2}+8} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{\pi \sqrt{2}}{64}\).

Step 1: Identify the form of the expression

Our integrand is \(\frac{1}{16x^2 + 8}\). We can factor out the 8 in the denominator to identify the form which is similar to \(a^2 + b^2x^2\). Rewriting gives us: \[\frac{1}{8(1 + 2x^2)}\], so here \(a^2 = 1\) and \(b^2 = 2\).

Step 2: Choose the appropriate trigonometric substitution

Since our expression is of the form \(a^2 + b^2x^2\), we use the substitution \(x = (a/b) \tan(\theta)\). Thus, \(x = \frac{1}{\sqrt{2}} \tan(\theta)\).

Step 3: Find the differential dx

Differentiate the substitution: if \(x = \frac{1}{\sqrt{2}} \tan(\theta)\), then \(dx = \frac{1}{\sqrt{2}} \sec^2(\theta) d\theta\).

Step 4: Convert limits of integration

When \(x = 0\), \(\tan(\theta) = 0\), so \(\theta = 0\). When \(x = \frac{1}{\sqrt{2}}\), \(\tan(\theta) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1\), so \(\theta = \frac{\pi}{4}\).

Step 5: Substitute and simplify the integral

Substitute back into the integral: \[\int_{0}^{\pi/4} \frac{1}{8(1 + 2(\frac{1}{\sqrt{2}} \tan(\theta))^2)} \cdot \frac{1}{\sqrt{2}} \sec^2(\theta) d\theta\]. Simplifying the expression yields: \[\int_{0}^{\pi/4} \frac{1}{8} \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sec^2(\theta)} \cdot \sec^2(\theta) d\theta = \frac{1}{8\sqrt{2}} \int_{0}^{\pi/4} d\theta\].

Step 6: Evaluate the integral

The integral \(\frac{1}{8\sqrt{2}} \int_{0}^{\pi/4} d\theta\) becomes \(\frac{1}{8\sqrt{2}} \cdot \left[ \theta \right]_{0}^{\pi/4}\). Evaluate this as \(\frac{1}{8\sqrt{2}}\left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{32\sqrt{2}}\).

Step 7: Simplify the result into a more familiar form

Recognize that \(\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\), so \(\frac{\pi}{32\sqrt{2}} = \frac{\pi}{32} \cdot \frac{\sqrt{2}}{2} = \frac{\pi\sqrt{2}}{64}\).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of calculus that focuses on accumulation of quantities and the areas under curves. In simple terms, while differential calculus deals with rates of change, integral calculus is about sums or totals.

Integrals come in two main types: indefinite integrals, which represent antiderivatives, and definite integrals, which calculate the area under a curve between two boundaries. In this exercise, we're dealing with a definite integral. Our task is to find the region under the curve of \( \frac{1}{16x^2 + 8} \) from \( x = 0 \) to \( x = \frac{1}{\sqrt{2}} \).

Integral calculus often involves techniques like substitution to make the integration process easier, especially when direct integration appears complex. By using trigonometric identities and substitution methods, we simplify integrals into more manageable forms.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions like sine, cosine, and tangent, fundamental to trigonometry. These identities help simplify complex mathematical expressions, particularly in integral calculus.

In this exercise, a trigonometric substitution was used, specifically relating to the expression \( a^2 + b^2x^2 \). We replaced \( x \) with \( (a/b) \tan(\theta) \), leveraging the identity \( 1 + \tan^2(\theta) = \sec^2(\theta) \). This substitution allowed us to simplify the integral significantly.

Trigonometric identities help transform tricky algebraic expressions into trigonometric ones, making them more straightforward to integrate once put into standard trigonometric form.

Definite Integrals

Definite integrals are used to evaluate the accumulation of a quantity over an interval, effectively finding the area under a curve from one point to another on the x-axis. In definite integrals, limits of integration define the range of interest.

In this exercise, we evaluated the integral from \( 0 \) to \( \frac{1}{\sqrt{2}} \), computing the net area under the curve. This involved transforming both the limits and the integrand into terms of \( \theta \) due to the trigonometric substitution. When we transformed the limits, \( x = 0 \) converted \( \theta = 0 \) and \( x = \frac{1}{\sqrt{2}} \) became \( \theta = \frac{\pi}{4} \).

Evaluating the definite integral involved calculating the integral expression using these transformed limits, ultimately yielding the accumulated value between these two boundaries.

Integration Techniques

Using integration techniques like substitution can vastly simplify the process of evaluating integrals. There are several techniques available, and choosing the right one often depends on recognizing the form of the integrand.

In this exercise, we employed "trigonometric substitution," a method deployed when integrands involve terms like \( a^2 + b^2x^2 \). By substituting \( x \) with a trigonometric function, we expressed the integral in a trigonometric context, making it easier to integrate.

The method used here turned a potentially cumbersome integration task into a simple one by transforming \( x = \frac{1}{\sqrt{2}}\tan(\theta) \). This allowed us to apply standard trigonometric identities to further simplify the expression and then straightforwardly integrate it over the new range of \( \theta \). Integrals that initially appear complex can often be simplified through the clever application of trigonometric identities and substitutions.