Problem 85 - Calculus Single Variable

Question

Suppose that \(a\) is a positive constant. Show that the substitution \(x=a \sin (\theta)\) leads to the formula $$ \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{a} \ln \left(\frac{x+a}{\sqrt{a^{2}-x^{2}}}\right)+C $$ Show that the substitution \(x=a \sec (\theta)\) leads to the formula $$ \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{a} \ln \left(\frac{x+a}{\sqrt{x^{2}-a^{2}}}\right)+C $$ Use these formulas to evaluate \(\alpha=\int_{0}^{1 / 2} \frac{1}{1-x^{2}} d x\) and \(\beta=\int_{\sqrt{2}}^{2} \frac{1}{1-x^{2}} d x .\) Show that the identity $$ \frac{1}{a^{2}-x^{2}}=\frac{1}{2 a}\left(\frac{1}{x+a}-\frac{1}{x-a}\right) $$ leads to the formula $$ \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \ln \left(\left|\frac{x+a}{x-a}\right|\right)+C $$ and use it to verify your answers for \(\alpha\) and \(\beta\).

Step-by-Step Solution

Answer

α = ln(2), β = ln(1 + √2). Verified with formula: ln|\(\frac{x+a}{x-a}\)|.

Step 1: Trigonometric Substitution x = a sin(θ)

Start with the substitution \(x = a \sin(\theta)\). Then, \( dx = a \cos(\theta) d\theta \). Substitute into the integral:\[ \int \frac{1}{a^2 - x^2} \, dx = \int \frac{1}{a^2 - (a^2 \sin^2(\theta))} \, a \cos(\theta) d\theta \]\( = \int \frac{a \cos(\theta)}{a^2 \cos^2(\theta)} \, d\theta \)\( = \int \frac{1}{a \cos(\theta)} \, d\theta \)\( = \frac{1}{a} \int \sec(\theta) \, d\theta \) where \(\sec(\theta) = \frac{1}{\cos(\theta)} \).

Step 2: Evaluate the Integral with x = a sin(θ)

The integral of \(\sec(\theta)\) is \(\ln|\sec(\theta) + \tan(\theta)| + C\). Thus:\[ \frac{1}{a} \ln|\sec(\theta) + \tan(\theta)| + C \]Convert back to terms of \(x\):\( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{a}{\sqrt{a^2 - x^2}}\) and \( \tan(\theta) = \frac{x}{\sqrt{a^2 - x^2}} \).Thus, the expression becomes:\[ \frac{1}{a} \ln \left(\frac{x + a}{\sqrt{a^2 - x^2}}\right) + C \]

Step 3: Trigonometric Substitution x = a sec(θ)

Use the substitution \(x = a \sec(\theta)\), then \(dx = a \sec(\theta) \tan(\theta) \, d\theta\).Substitute into the original integral:\[ \int \frac{1}{a^2-x^2} \, dx = \int \frac{1}{a^2 - a^2 \sec^2(\theta)} \cdot a \sec(\theta) \tan(\theta) \, d\theta \]\( = \int \frac{-a \sec(\theta) \tan(\theta)}{a^2 \tan^2(\theta)} \, d\theta \)\( = -\frac{1}{a} \int \csc(\theta) \, d\theta \) where \( \csc(\theta) = \frac{1}{\sin(\theta)} \).

Step 4: Evaluate the Integral with x = a sec(θ)

The integral of \(\csc(\theta)\) is \(-\ln|\csc(\theta) + \cot(\theta)| + C\). Thus:\[ -\frac{1}{a} \ln|\csc(\theta) + \cot(\theta)| + C \]In terms of \(x\):- \(\csc(\theta) = \frac{x}{\sqrt{x^2 - a^2}}\)- \(\cot(\theta) = \frac{a}{\sqrt{x^2-a^2}}\)Become:\[ \frac{1}{a} \ln \left(\frac{x+a}{\sqrt{x^2-a^2}}\right) + C \]

Step 5: Evaluate α and β

For \(\alpha = \int_0^{1/2} \frac{1}{1-x^2} \, dx\), use \(a = 1\) for the first formula:\[ \left. \frac{1}{1} \ln \left(\frac{x+1}{\sqrt{1 - x^2}}\right) \right|_0^{1/2} \]Calculating, you get:\(\alpha = \ln(2)\)For \(\beta = \int_{\sqrt{2}}^2 \frac{1}{1-x^2} \, dx\), use \(a = 1\) for the second formula:\[ \left. \frac{1}{1} \ln \left(\frac{x+1}{\sqrt{x^2-1}}\right) \right|_{\sqrt{2}}^{2} \]Calculating, you get:\(\beta = \ln \left(1 + \sqrt{2} \right)\)

Step 6: Verify Identity

Start with the identity: \( \frac{1}{a^{2}-x^{2}}=\frac{1}{2 a}\left(\frac{1}{x+a}-\frac{1}{x-a}\right) \)Integrate:\[ \frac{1}{2a} \left( \ln|x+a| - \ln|x-a| \right) + C \]This simplifies to:\[ \frac{1}{2a} \ln \left| \frac{x+a}{x-a} \right| + C \]

Step 7: Verify α and β with Identity

Use the verified formula for \(\alpha\) and \(\beta\):For \(\alpha\):\[ \int_0^{1/2} \frac{1}{1-x^2} \, dx = \frac{1}{2} \ln \left| \frac{1/2+1}{1/2-1} \right| \]Calculating gives \(\ln(2)\) as expected.For \(\beta\):\[ \int_{\sqrt{2}}^2 \frac{1}{1-x^2} \, dx = \frac{1}{2} \ln \left| \frac{2+1}{\sqrt{2}+1} \right| \]Calculating gives \(\ln(1+\sqrt{2})\) confirming previous results.

Key Concepts

Integration Techniques

Integration techniques are powerful tools for solving complex integral problems. One popular method is trigonometric substitution. This technique is particularly useful when dealing with integrals involving the square root of expressions like \(a^2 - x^2\) or \(x^2 - a^2\). Trigonometric substitution leverages trigonometric identities to simplify these expressions. For instance, substituting \(x = a\sin(\theta)\) transforms \(x^2\) into \(a^2\sin^2(\theta)\), simplifying the integral considerably. This substitution helps by converting the problem into one involving trigonometric functions, which are easier to integrate or differentiate.Another substitution, such as \(x = a\sec(\theta)\), applies to integrals with terms like \(\sqrt{x^2 - a^2}\). By substituting, you transform these expressions into more familiar trigonometric forms that can be managed with standard calculus techniques. The purpose of these substitutions is to reduce the problem to one involving basic integrals that have well-known solutions. It is essential to revert back to terms of \(x\) after performing the integration to ensure it aligns well with the original problem context. This complete cycle from substitution to integration and finally to substitution back connects complex mathematical expressions to simple trigonometric relationships.

Definite Integrals

Definite integrals represent the area under a curve between two limits, providing a numerical value rather than a general formula with a constant of integration. They are vital in calculating accumulated quantities, such as distance, area, or volume. To solve definite integrals involving trigonometric substitutions, it is often necessary to adjust the boundaries of integration to match the substitution. For example, when substituting \(x = a\sin(\theta)\), the limits for \(x\) must be translated to corresponding \(\theta\) values to maintain logical continuity in evaluation.After integrating using the trigonometric substitution, the resulting expression often includes trigonometric functions. These functions have to be evaluated at the substituted bounds. Finally, these results are used to compute the definite integral by subtracting the lower limit value from the upper one. This process brings out the precise value for the definite integral, offering insights into the measure across the specified interval.By thorough usage of trigonometric substitutions and careful tracking of limits, we can simplify the integral into more handleable forms and effectively find the area or value represented by the definite integral.

Trigonometric Identities

Trigonometric identities are fundamental relationships between trigonometric functions. They serve as essential tools to simplify integrals involving complex trigonometric expressions. Key identities include the Pythagorean identities, such as \(\sin^2(\theta) + \cos^2(\theta) = 1\), which are crucial for simplifying expressions under square roots in integrals. Using identities like \(\sec(\theta) = \frac{1}{\cos(\theta)}\) or \(\csc(\theta) = \frac{1}{\sin(\theta)}\), we can transform difficult integrals into more manageable forms.In trigonometric substitution, these identities help rewrite integrals to utilize simpler, more direct integration techniques. For instance, the use of \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\) can transform quotients of sine and cosine into products or sums. Often, knowing the inverse relationships, like \(\theta = \arcsin\left(\frac{x}{a}\right)\), is essential for reverting initial substitutions back to terms of \(x\) after integration is complete.Understanding and applying trigonometric identities enhance the flexibility and power of integral calculus. They are especially useful when verifying solutions, as they allow us to manipulate expressions into equivalent and often simpler forms. Identities enable mathematicians and students alike to approach integrals with confidence, knowing they have a toolbox of transformations at their disposal.