Problem 57 - Calculus Early Transcendentals

Question

Determine the following indefinite integrals. $$\int \frac{d x}{x \sqrt{4-x^{8}}}$$

Step-by-Step Solution

Answer

Question: Determine the indefinite integral of the function \(\int \frac{d x}{x \sqrt{4-x^{8}}}\). Answer: The indefinite integral of \(\int \frac{d x}{x \sqrt{4-x^{8}}}\) is \(4\sin^{-1}{\frac{x^4}{2}} + C\).

Step 1: Identify the Substitution

For this integral, let's use the substitution \(x^{4} = 2u\). Then, $$x^{8} = 4u^2$$ and $$4\,dx = 8u\,du$$.

Step 2: Rewrite the Integral Using the Substitution

Using the substitution, we can rewrite the integral as follows: $$\int \frac{d x}{x \sqrt{4-x^{8}}} = \int \frac{8u}{(2u) \sqrt{4-4u^2}}\,du$$

Step 3: Simplify the Integral

Now, we have $$\int \frac{8u}{(2u) \sqrt{4-4u^2}}\,du = \int \frac{4}{\sqrt{1-u^2}}\,du$$

Step 4: Identify the Trigonometric Substitution

We notice that the integral is of the form \(\frac{C}{\sqrt{1-x^2}}\), which reminds us of the sine function's derivative. Let's make another substitution: $$u = \sin{v}\quad\Longrightarrow\quad du = \cos{v}\,dv$$

Step 5: Rewrite the Integral Using the Trigonometric Substitution

Substituting \(u\) with \(\sin{v}\), we have: $$\int \frac{4}{\sqrt{1-u^2}}\,du = \int \frac{4}{\sqrt{1-\sin^2{v}}}\cos{v}\,dv$$

Step 6: Simplify the Integral

We know that \(\sqrt{1-\sin^2{v}} = \cos{v}\), so we have: $$\int \frac{4}{\sqrt{1-\sin^2{v}}}\cos{v}\,dv = \int 4\,dv$$

Step 7: Integrate the Simplified Integral

Integrating with respect to \(v\), we get: $$\int 4\,dv = 4v + C$$

Step 8: Perform Back Substitution

Now, we backtrack through our substitutions. First, we have \(v = \sin^{-1}{u}\), and second, we have \(u = \frac{x^4}{2}\). Therefore: $$4v + C = 4\sin^{-1}{\frac{x^4}{2}} + C$$

Step 9: Write the Final Answer

Thus, the indefinite integral is: $$\int \frac{d x}{x \sqrt{4-x^{8}}} = 4\sin^{-1}{\frac{x^4}{2}} + C$$

Key Concepts

Substitution Method

The substitution method is a powerful technique in calculus, particularly for finding indefinite integrals. The goal is to simplify the integral by introducing a new variable. This can help when the integral involves complex expressions.
To apply this method, we first choose a substitution that reduces the complexity of the equation. For example, in the given exercise, we use the substitution \(x^{4} = 2u\).
This choice modifies the parameters of the integral into a more manageable form.Key steps include:

• Identify the substitution: Choose an expression within the integral that can be replaced.
• Rewrite in terms of the new variable: Replace the original variables with your substitution.
• Change all expressions including \(dx\), transforming them entirely into the new variable domain.

After simplifying the integral with substitution, you execute the integration and then back-substitute to express the result in the original variable form.

Trigonometric Substitution

Trigonometric substitution is particularly useful when dealing with integrals containing expressions of square roots of quadratic forms. It involves introducing a trigonometric function to simplify the integral’s integrand.
In this exercise, after the initial substitution, the integrand became reminiscent of the derivative form of trigonometric functions like sine and cosine.The process involves:

• Recognize the trigonometric identity: Identify segments of the integrand that resemble trigonometric forms, such as \(\sqrt{1-u^2}\), which hints at \(\sin u\).
• Make the trigonometric substitution: Here, it's helpful to substitute \(u = \sin v\).
• Differentiate to find \(du\): If \(u = \sin v\), then \(du = \cos v\,dv\).

Substituting back, you can often transform the integral into a basic form that is much simpler to integrate.

Integral Simplification

Integral simplification often follows after using substitution methods. The transformed integral might initially still appear complicated. Thus we simplify further to make integration easier.
In our exercise, once the trigonometric substitution was applied, the expression \(\sqrt{1- ext{something}^2}\) naturally reduces, using trigonometric identities.Steps in simplification include:

• Simplify square root expressions: Use trigonometric identities like \(\cos v\) for \(\sqrt{1- ext{sin}^2{v}}\).
• Cancel out terms: Simplify division or multiplication if trigonometric substitutions align well.
• Integrate simpler expressions: Identify parts of the integral that are elementary to compute directly.

In this problem, simplifying \(\frac{4}{\sqrt{1-u^2}}\cos v\) to \(4\) allowed straightforward integration.

Inverse Trigonometric Functions

Inverse trigonometric functions appear often in calculus, especially when solving integrals involving trigonometric identities or solutions that require back-substitution. They provide us a way to express certain integrals in terms of angles.With our given problem, the final solution depended on the inverse sine function. From the trigonometric substitution step:

• Identify inverse function relationships: When performing substitutions back, recognize which inverse trigonometric relationships apply, such as \(v = \sin^{-1}{u}\).
• Use these relationships to express the anti-derivative: The simplified integral \(\int 4\,dv\) led to \(4v\) which corresponds back to the variable \(u\) using inverse functions.

Ultimately, the integral of \(4\) with respect to \(v\) yields \(4 \sin^{-1}{\left(\frac{x^4}{2}\right)} + C\), illustrating how these inverse functions can complete the integration process.