Problem 393 - Calculus Volume 1

Question

In the following exercises, evaluate each integral in terms of an inverse trigonometric function. $$\int_{\sqrt{3}}^{1} \frac{d x}{\sqrt{1+x^{2}}}$$

Step-by-Step Solution

Answer

\( \frac{1}{2} \ln \left(\frac{2 + \sqrt{3}}{3 + \sqrt{8}}\right) \)

Step 1: Identify the appropriate inverse trigonometric function

The integral \( \int \frac{dx}{\sqrt{1+x^2}} \) is the standard form for the derivative of the inverse hyperbolic sine function. Recognizing this form allows us to utilize the inverse trigonometric function \( \sinh^{-1}(x) \).

Step 2: Apply the antiderivative

Integrate the function with respect to \( x \) to obtain \( \sinh^{-1}(x) + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

Use the limits of integration to evaluate \( \left. \sinh^{-1}(x) \right|_{\sqrt{3}}^1 \), which translates to \( \sinh^{-1}(1) - \sinh^{-1}(\sqrt{3}) \).

Step 4: Evaluate the inverse hyperbolic sine values

Calculate \( \sinh^{-1}(1) \) which is equivalent to the value \( \frac{\ln(2 + \sqrt{3})}{2} \), and \( \sinh^{-1}(\sqrt{3}) \) which corresponds to \( \frac{\ln(3 + \sqrt{8})}{2} \).

Step 5: Simplify the final result

Subtract the evaluated values: \[ \frac{\ln(2 + \sqrt{3})}{2} - \frac{\ln(3 + \sqrt{8})}{2} = \frac{\ln(2 + \sqrt{3}) - \ln(3 + \sqrt{8})}{2}. \] Therefore, the solution simplifies to \( \frac{1}{2} \ln \left(\frac{2 + \sqrt{3}}{3 + \sqrt{8}}\right) \).

Key Concepts

Integrals

Integrals are a fundamental tool in calculus, helping us to find areas under curves, among many other applications. The integral of a function tries to capture the accumulation of quantities, like finding the total distance traveled given a velocity function.

When evaluating an integral like \( \int \frac{dx}{\sqrt{1+x^2}} \), you're identifying patterns or standard forms that correspond to known functions. Recognizing these forms allows us to apply easier methods rather than deriving results from scratch every time.

• The specific form \( \int \frac{dx}{\sqrt{1+x^2}} \) is notable because it matches a derivative of an inverse hyperbolic function.
• Using these patterns streamlines solving, especially for complex functions.

Once you identify these forms, you can proceed quickly to evaluation, using antiderivatives and constants of integration effectively.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are analogs of inverse trigonometric functions that often arise in calculus, especially when dealing with integrals involving square roots.

For instance, the inverse hyperbolic sine function, \( \sinh^{-1}(x) \), which is used in our problem, comes from the expression \( \int \frac{dx}{\sqrt{1+x^2}} \). This is a typical form seen when encountering integrals of this nature.

• They provide solutions for integrals that have square roots with sums inside, not differences.
• Understanding the link between these functions helps demystify why certain integrals simplify the way they do.
• Solving these integrals involves using identities and properties specific to hyperbolic functions.

By understanding inverse hyperbolic functions, you gain a powerful tool in calculus that makes solving certain problems much more manageable.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes a connection between differentiation and integration. It's crucial in evaluating definite integrals, like the one given in our exercise.

The theorem states that if \( F \) is an antiderivative of \( f \) over an interval, then the integral of \( f \) from \( a \) to \( b \) is \( F(b) - F(a) \). This means you can evaluate an integral by merely finding its antiderivative and applying the limits of integration.

• It simplifies the problem-solving process by connecting the concept of area under a curve with the slope of a tangent.
• By recognizing the function's antiderivative, you can directly evaluate the integral's endpoints.
• In our solution, \( \left. \sinh^{-1}(x) \right|_{\sqrt{3}}^1 \) precisely showcases this concept, translating areas into manageable subtraction between two points.

Using this theorem lets us efficiently handle integrals, focusing on finding antiderivatives and substituting values, thus simplifying what might seem an intimidating task.