Problem 44 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(\int \sqrt{k^{2}-x^{2}} d x\) for any constant \(k\).

Step-by-Step Solution

Answer

\(\int \sqrt{k^{2}-x^{2}} d x = \frac{k}{2} \arcsin(\frac{x}{k}) + \frac{x}{2} \sqrt{k^{2}-x^{2}} + C \)

Step 1: Set Up the Trigonometric Substitution

Choose the substitution \( x = k \sin(\theta) \), because this makes the integral simpler as \( \sqrt{k^{2}-x^{2}} = \sqrt{k^{2}-k^{2} \sin^{2}(\theta)} = k \cos(\theta) \). Also, \( dx = k \cos(\theta) d\theta \).

Step 2: Substitute

Substitute the identities from Step 1 into the integral, it becomes \( \int k \cos^{2}(\theta) d\theta \). The integral simplifies because the complex square root is replaced with a simple trigonometric function.

Step 3: Use the Double-Angle Formula

We can simplify the integrand more by applying the double-angle formula for cos, which expresses \( \cos^{2}(\theta) \) as \( \frac{1+\cos(2\theta)}{2} \). So the integral becomes \( \int k \frac{1+\cos(2\theta)}{2} d\theta \), effectively lowering the power of cos from 2 to 1.

Step 4: Distribute and Integrate

Distribute \( k \) inside the integral and separate it into two integrals. The integral becomes \( \frac{k}{2}\int d\theta + \frac{k}{2}\int \cos(2\theta) d\theta \). The first integral simply becomes \( \theta \), and the second integral can be solved using the reverse chain rule, with the solution being \( \frac{1}{2}\sin(2\theta) \)

Step 5: Substitute back to x

Replace \( \theta \) with \( \arcsin(\frac{x}{k}) \) and \( \sin(2\theta) \) with \( 2\sin(\theta) \cos(\theta) \) to express the answer in terms of \( x \). Using the identity \( \cos(\theta) = \sqrt{1-\sin^{2}(\theta)} = \sqrt{1-(\frac{x}{k})^{2}} \), you obtain the final answer: \( \frac{k}{2} \arcsin(\frac{x}{k}) + \frac{x}{2} \sqrt{k^{2}-x^{2}} \), with an integration constant \( C \) added at the end.

Key Concepts

Integration Techniques

Integration, especially with functions involving radicals, can be tricky. One powerful tool at your disposal is **Trigonometric Substitution**.
This method leverages trigonometric identities to simplify the integral, which makes it much easier to solve.
Here's how it works when faced with the function \( \int \sqrt{k^{2}-x^{2}} \, dx \). First, notice the expression under the radical \( k^2 - x^2 \). It resembles \( \sin^2 \theta \) in the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). This gives us a clue to use a substitution like \( x = k \sin \theta \).
By making this substitution, the integral simplifies because the square root term turns into \( k \cos \theta \). This eliminates the complexity introduced by the radical, helping us focus on integrating a trigonometric function.
Trigonometric substitution is just one among many integration techniques, but it's particularly invaluable when dealing with integrals that include \( \sqrt{a^2 - x^2} \), \( \sqrt{x^2 - a^2} \), or \( \sqrt{x^2 + a^2} \). Each of these forms can be tackled by an appropriate trigonometric identity, making them a breeze to integrate once the problem is set up correctly.

Definite Integrals

Definite integrals help us calculate the area under a curve between two specified limits. However, the integral in the exercise \( \int \sqrt{k^{2}-x^{2}} \, dx \) is an indefinite integral because it doesn't have upper and lower bounds.
While this exercise focuses on indefinite integration, it's always helpful to think about how the process might differ. With a definite integral, after solving the antiderivative, you need to evaluate it at the given limits to find the actual area.
When dealing with definite integrals in the context of trigonometric substitution, don't forget to change the limits of integration to correspond with your substitution.

• Ensure any substitution is reverted back to raw \( x \)-terms if required, especially for final evaluation.
• The addition of limits transforms the antiderivative into a specific numerical answer representing the area.

Understanding this helps connect the processes of solving indefinite integrals and then further applying them to solve definite integrals, fortifying your calculus toolbox.

Trigonometric Functions

Trigonometric functions like sine and cosine are more than just triangle ratios – they're vital tools in calculus. In the exercise, we use sine and cosine to simplify an integral by employing their relationships within trigonometric identities.
Here’s how they come into play:
When we substitute \( x = k \sin \theta \), we directly tie \( x \) to the sine function. This aligns with \( \sin^2 \theta + \cos^2 \theta = 1 \) to simplify \( \sqrt{k^2 - x^2} \), reducing it to \( k \cos \theta \).
Further simplification uses **trigonometric identities** such as the double-angle formula \( \cos^2 \theta = \frac{1+\cos(2\theta)}{2} \). This change reduces the powers of cosine to more manageable terms in the integral.By being familiar with these trigonometric functions and their identities:

• You can convert complex radical expressions into simpler trigonometric forms.
• You can connect calculus to pre-calculus concepts, like trigonometric identities, making integration a more approachable task.

By mastering these connections, integrals involving radicals become not just solvable but a compelling exercise in mathematical relationships.