Problem 29 - Calculus Single Variable

Question

Evaluate the given integral by making a trigonometric substitution (even if you spot another way to evaluate the integral). $$ \int_{1}^{\sqrt{2}} \sqrt{x^{2}-1} d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( 1 \).

Step 1: Identify the Trigonometric Substitution

The integral involves the expression \( \sqrt{x^2 - 1} \), which suggests the hyperbolic substitution \( x = \sec(\theta) \), as \( \sqrt{(\sec(\theta))^2 - 1} = \tan(\theta) \). Therefore, set \( x = \sec(\theta) \).

Step 2: Find the Differential

Differentiate \( x = \sec(\theta) \) with respect to \( \theta \). We get \( dx = \sec(\theta)\tan(\theta) \, d\theta \).

Step 3: Change the Limits of Integration

Change the limits of integration based on the substitution \( x = \sec(\theta) \). When \( x = 1 \), \( \sec(\theta) = 1 \) implies \( \theta = 0 \). When \( x = \sqrt{2} \), \( \sec(\theta) = \sqrt{2} \) implies \( \theta = \frac{\pi}{4} \).

Step 4: Substitute into the Integral

Substitute \( x = \sec(\theta) \) and \( dx = \sec(\theta)\tan(\theta) \, d\theta \) into the integral. This gives:\[\int_{0}^{\frac{\pi}{4}} \tan(\theta) \cdot \sec(\theta)\tan(\theta) \, d\theta = \int_{0}^{\frac{\pi}{4}} \sec(\theta)\tan^2(\theta) \, d\theta.\]

Step 5: Simplify the Integral

Recall the identity \( \tan^2(\theta) = \sec^2(\theta) - 1 \) to simplify the integral:\[\int_{0}^{\frac{\pi}{4}} \sec(\theta)(\sec^2(\theta) - 1) \, d\theta = \int_{0}^{\frac{\pi}{4}} (\sec^3(\theta) - \sec(\theta)) \, d\theta.\]

Step 6: Evaluate Simplified Integral

Split the integral into two parts, \( \int_{0}^{\frac{\pi}{4}} \sec^3(\theta) \, d\theta - \int_{0}^{\frac{\pi}{4}} \sec(\theta) \, d\theta \). Evaluate both parts separately, often using integration by parts for the \( \sec^3(\theta) \) term and the basic integration formula for \( \sec(\theta) \).

Step 7: Substitute Back the Original Variable

After finding the solution to the integral in terms of \( \theta \), convert back to the variable \( x \) using \( x = \sec(\theta) \).

Step 8: Calculate Final Result

Calculate the final value by plugging the new limits back into the function obtained from the previous evaluations.

Key Concepts

Hyperbolic Functions

Hyperbolic functions often appear in solving integrals involving expressions like \( \sqrt{x^2 - 1} \). These functions, similar to trigonometric ones, are derived from hyperbolas. The primary hyperbolic functions include \( \sinh \), \( \cosh \), and \( \tanh \);these functions relate to the analogs of sine, cosine, and tangent in trigonometry, respectively.

In trigonometric substitution, hyperbolic functions can simplify integrals involving radicals due to their inherent properties. For instance, consider \( x = \sec(\theta) \), as used in the solution. This substitution provides a neat manipulation where \( \sqrt{x^2 - 1} = \tan(\theta) \). It helps to transform the integral into one that involves hyperbolic properties, simplifying the problem.

• \( \sinh = \frac{e^x - e^{-x}}{2} \), relates to the hyperbolic sine function.
• \( \cosh = \frac{e^x + e^{-x}}{2} \), relates to hyperbolic cosine.
• These functions are particularly useful when variables have large exponents or roots.

By understanding and utilizing hyperbolic functions, we have an efficient tool in transforming and evaluating complex integrals, like the one shown in the exercise.

Integration by Parts

Integration by parts is a technique often used to handle integrals involving the product of functions. It's similar to the product rule for differentiation and is especially useful for functions that include products of algebra and trigonometry, such as our \( \sec^3(\theta) \) term in the exercise.

When applying integration by parts, we start with the formula: \[\int u \, dv = uv - \int v \, du\] - **Choosing \( u \) and \( dv \):** It's crucial to choose parts of the integral wisely. Typically, we let \( u \) be a function that simplifies when differentiated and \( dv \) as something straightforward to integrate.- **Our Example as Used:** For \( \sec^3(\theta) \), we set \( u = \sec(\theta) \) and \( dv = \sec^2(\theta) \, d\theta \). This choice simplifies the integral because it's easier to differentiate \( \sec(\theta) \) and integrate \( \sec^2(\theta) \).By applying integration by parts, we simplify and tackle parts of the integrals that are otherwise complex, making it easier to integrate stepwise as seen in our solution.

Trigonometric Identities

Trigonometric identities are powerful tools used in algebra to simplify expressions and solve equations, especially those involved with trigonometric substitution like we performed. They help break down complex expressions into manageable forms.

For the integral in the exercise, we used the identity:\[\tan^2(\theta) = \sec^2(\theta) - 1\]This identity transformed the integral into a more straightforward form, \( \sec^3(\theta) - \sec(\theta) \), allowing further simplification.

• **Double Angle Identities** remove angles in trigonometric functions.
• **Pythagorean Identities** allow conversion between squares of sine, cosine, and tangent functions.
• These identities facilitate the manipulation and simplification of trigonometric expressions.

By using identities, we can see how expressions reduce significantly, making integration much more feasible. This is imperative for assessing integrals like the one in this exercise.

Definite Integrals

Definite integrals compute the exact value of the area under a curve within specified limits. It's a cornerstone concept in calculus used for finding accumulated quantities over intervals. In our exercise, the limits were \( 1 \) and \( \sqrt{2} \), converted into \( \theta \) using substitution.

To deal with definite integrals:- **Change of Limits:** As you substitute variables, change your limits accordingly. When \( x = 1 \), \( \theta = 0 \); when \( x = \sqrt{2} \), \( \theta = \frac{\pi}{4} \).- **Evaluate the Result:** After integration, plug the limits into the resulting expressions to find the specific numerical areas or values. For the integral involving \( \sec^3(\theta) \), we calculated from \( 0 \) to \( \frac{\pi}{4} \).Definite integrals provide not only the means to calculate areas but also the methodology to transform complex problems into solvable ones. They allow for the transitioning back into original variables after solving in terms of substituted ones. By learning how to manage limits and changes in variables, you develop strong skills for evaluating definite integrals correctly.