Problem 20 - Calculus: One and Several Variables

Question

Calculate. $$\int \frac{d x}{x^{2} \sqrt{a^{2}-x^{2}}}$$.

Step-by-Step Solution

Answer

The short answer for calculating the integral $$\int \frac{d x}{x^{2} \sqrt{a^{2}-x^{2}}}$$ is: \[-\frac{\sqrt{a^2-x^2}}{a^2x} + C\]

Step 1: Trigonometric Substitution

Make the substitution \(x = a\sin{\theta}\), where \(a\) is a constant. Then, we have \(dx = a\cos{\theta} d\theta\), and we can rewrite the given integral in terms of \(\theta\): \[\int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} \, dx = \int \frac{a\cos{\theta}}{(a\sin{\theta})^{2} \sqrt{a^{2} - (a\sin{\theta})^{2}}} \, d\theta\]

Step 2: Simplify the integral

Now, we simplify the expression inside the integral: \[\int \frac{a\cos{\theta}}{a^{2}\sin^{2}{\theta} \sqrt{a^{2} - a^{2}\sin^{2}{\theta}}} \, d\theta\] Using the identity \(\cos^{2}{\theta} = 1 - \sin^{2}{\theta}\), we can rewrite the square root term as: \[\sqrt{a^{2} - a^{2}\sin^{2}{\theta}} = \sqrt{a^{2}\cos^{2}{\theta}} = a \cos{\theta}\] Then, the integral becomes: \[\int \frac{a\cos{\theta}}{a^{2}\sin^{2}{\theta} \cdot a \cos{\theta}} \, d\theta = \int \frac{1}{a^{2}\sin^{2}{\theta}} \, d\theta\]

Step 3: Integrate

Now, we integrate the simplified integral: \[\int \frac{1}{a^{2}\sin^{2}{\theta}} \, d\theta = \frac{1}{a^{2}} \int \csc^{2}{\theta} \, d\theta\] Recall that the integral of \(\csc^{2}{\theta}\) is \(-\cot{\theta}\). So, we have: \[-\frac{1}{a^{2}} \cot{\theta} + C\]

Step 4: Convert back to x

Now, we convert our result back into terms of \(x\): First, from the substitution \(x = a\sin{\theta}\), we can find \(\cot{\theta}\) as follows: \[\sin \theta= \frac{x}{a} \Rightarrow \cos{\theta} = \sqrt{1-\sin^{2}{\theta}} = \frac{\sqrt{a^{2} - x^{2}}}{a}\] Therefore, \(\cot{\theta} = \frac{\cos{\theta}}{\sin{\theta}} = \frac{\sqrt{a^2-x^2}}{x}\), and we get the final result: \[-\frac{1}{a^{2}} \left(\frac{\sqrt{a^2-x^2}}{x}\right) + C = -\frac{\sqrt{a^2-x^2}}{a^2x} + C\]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a strategy used in integral calculus to simplify the integration of expressions involving square roots. This technique is especially useful when dealing with integrals containing expressions like \(\sqrt{a^2 - x^2}\). The goal here is to make the integral easier to evaluate by substituting a trigonometric function for the variable.

In this particular problem, the substitution \(x = a\sin\theta\) is made. This choice is beneficial because it uses the identity \(1 - \sin^2\theta = \cos^2\theta\), which simplifies the expression inside the square root to \(a\cos\theta\). By also substituting \(dx = a\cos\theta\,d\theta\), the original integral transforms completely in terms of \(\theta\), making it more straightforward to handle.

This method is advantageous because it reduces a complex algebraic function into a more manageable trigonometric form. Remember that each trigonometric substitution aligns with a specific form of expression:

• Use \(x = a\sin\theta\) for \(\sqrt{a^2 - x^2}\).
• Use \(x = a\tan\theta\) for \(\sqrt{a^2 + x^2}\).
• Use \(x = a\sec\theta\) for \(\sqrt{x^2 - a^2}\).

Therefore, knowing which substitution to use, based on the form of the expression under the square root, is an essential skill in integral calculus.

Definite Integrals

Definite integrals represent the exact area under a curve between two specific bounds. They are an essential part of calculus, allowing us to calculate total quantities where variances exist over a certain interval. In contrast to indefinite integrals, which contain an arbitrary constant of integration \(C\), definite integrals evaluate to a number and do not include \(C\).

The process of solving definite integrals usually involves:

• Finding the antiderivative (indefinite integral) of the function.
• Evaluating this antiderivative at the upper and lower limits, followed by calculating the difference between these two values.

Although the original problem here is indefinite, being skilled in evaluating definite integrals means understanding limits \(a\) and \(b\), as well as potentially transforming these bounds when applying substitutions, like trigonometric ones.

In this learning context, understand that solving definite integrals might require sometimes the same substitution techniques you use for indefinite ones, ensuring you modify the limits when necessary to match the transformed variable.

Integration Techniques

Integration techniques refer to the various methods applied to solve integrals, whether definite or indefinite. These techniques simplify complex integrals, transforming them into interpretable forms. Mastering these allows you to tackle a broader range of problems. Some core techniques include:

• Substitution: As in the current problem with trigonometric substitution, it helps convert an intricate integral into a simpler form.
• Integration by Parts: Used particularly when the integrand is a product of two functions \(f(x)\) and \(g(x)\). It follows the formula: \(\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx\).
• Partial Fraction Decomposition: Where the integrand is a rational function, this technique breaks it down into simpler fractions making it easier to integrate.

Applying the right method speeds up the process and simplifies the work. Trigonometric substitution is just one tool in the larger toolkit of integration methods. Each technique has its uses, often stemming from the nature of the function inside the integral.

Brushing up on each method and practicing different problems will ensure you're confident and ready for a variety of integral-related challenges. Make sure to understand not just the calculation techniques, but also the underlying concepts guiding which technique to apply and when.