Problem 111 - Calculus Early Transcendentals

Question

Use the substitution \(u=x^{r}\) to show that \(\int \frac{d x}{x \sqrt{1-x^{2 r}}}=-\frac{1}{r} \operatorname{sech}^{-1} x^{r}+C,\) for \(r>0,\) and \(0

Step-by-Step Solution

Answer

Question: Find the integral of \(\int \frac{d x}{x \sqrt{1-x^{2r}}}\) using the substitution \(u = x^r\), where \(r > 0\) and \(0 < x < 1\). Answer: The integral equals to \(\displaystyle -\frac{1}{r} \operatorname{sech}^{-1} x^{r} + C\).

Step 1: Given Integral and Substitution

We are given the integral \(\int \frac{d x}{x \sqrt{1-x^{2r}}}\). We are asked to use the substitution \(u = x^r\).

Step 2: Substitute

Now we substitute \(u = x^r\) in the integral to make it easier to solve. \(\displaystyle \int \frac{d x}{x \sqrt{1-x^{2r}}} = \int \frac{d x}{x \sqrt{1-u^2}}\).

Step 3: Derivative of u

Next, we need to find the derivative of \(u\) with respect to \(x\) to make the substitution complete. \(\displaystyle \frac{du}{dx} = r \cdot x^{r-1}\). We can solve for \(dx\) from this expression: \(\displaystyle dx = \frac{1}{r \cdot x^{r-1}} du\).

Step 4: Substitute dx

Now we substitute \(\displaystyle dx = \frac{1}{r \cdot x^{r-1}} du\) into the integral: \(\displaystyle \int \frac{d x}{x \sqrt{1-u^2}} = \int \frac{\frac{1}{r \cdot x^{r-1}}du}{x \sqrt{1-u^2}}\).

Step 5: Simplify Integral

After substituting, we can simplify the expression: \(\displaystyle \int \frac{1}{r \cdot x^{r} \sqrt{1-u^2}}du\). Since \(u = x^r\), we can replace \(x^r\) with \(u\): \(\displaystyle \int \frac{1}{r \cdot u \sqrt{1-u^2}}du\).

Step 6: Solve the Integral

Now we need to solve the integral. \(\displaystyle -\frac{1}{r} \int \frac{du}{u\sqrt{1-u^2}}\). This is a standard integral with the expression \(\operatorname{sech}^{-1}(u)\) so, we can directly utilize the result: \(\displaystyle - \frac{1}{r} \operatorname{sech}^{-1}(u) + C\).

Step 7: Replace u with x^r

Finally, we will replace \(u\) with \(x^r\), our original substitution: \(\displaystyle - \frac{1}{r} \operatorname{sech}^{-1}(x^r) + C\). The integral \(\int \frac{d x}{x \sqrt{1-x^{2r}}}\) equals to \(\displaystyle -\frac{1}{r} \operatorname{sech}^{-1} x^{r} + C\), for \(r > 0\) and \(0 < x < 1\).

Key Concepts

Substitution Method

The substitution method, often known as "u-substitution," is a fundamental technique used in calculus for simplifying integrals. The idea is to introduce a new variable, typically called \( u \), which transforms the integral into a simpler form by substituting parts of the original expression. This method relies on recognizing a portion of the integration variable, typically expressed as a function, that can be replaced with this new variable.

In the exercise, the given substitution is \( u = x^r \). By doing this, the variable \( x \) is expressed in terms of \( u \), allowing the original integral \( \int \frac{d x}{x \sqrt{1-x^{2r}}} \) to be rewritten in an easier-to-integrate form. This substitution strategy simplifies the manipulations and calculations needed.

**Steps to Use the Substitution Method**

• Identify a portion of the integral that resembles the derivative of a function.
• Set \( u = f(x) \), where \( f(x) \) is a part of the integral.
• Calculate \( \frac{du}{dx} \) to determine \( dx \) in terms of \( du \).
• Rewrite the integral in terms of \( u \) and \( du \).
• Solve the new integral in terms of \( u \).
• Substitute back the original expression for \( u \) to find the solution in terms of \( x \).

This method is particularly helpful when dealing with definite integrals where limits can be easily transformed into the \( u \)-domain.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the inverses of hyperbolic functions, such as \( \sinh(x) \), \( \cosh(x) \), and \( \tanh(x) \). In the context of integrating functions, they offer solutions for integrals resembling certain forms.

In this exercise, the result of the integral involves the inverse hyperbolic secant function, \( \operatorname{sech}^{-1}(x) \). The expression \( \operatorname{sech}^{-1}(u) \) shows up after setting up a substitution and successfully transforming the integral into a known form.

**Why Inverse Hyperbolic Functions?**

• They can be used to simplify expressions, especially when integrals match their specific forms.
• Their derivatives lead to recognizably alternating forms, making them useful in solving complex integrals.

The transformation of integrals by recognizing the appearance of inverse hyperbolic functions often allows us to directly use known results, as performed in the given solution with \( \operatorname{sech}^{-1} \). This simplifies the process and aids in solving the integral with less computational effort.

Definite Integrals

Definite integrals calculate the net area under a curve between two specific limits. They are fundamental in evaluating total quantities, such as area, displacement, or any kind of aggregation based on continuous data.

While the example problem provided deals with an indefinite integral, it's helpful to understand how the solution can apply to definite integrals when limits are given. With a definite integral, after applying a substitution like \( u = x^r \), one must also transform the limits of integration accordingly. This step ensures calculations are exact and align with the integral's original bounds.

**Steps for Handling Definite Integrals with Substitution**

• First, express the function in terms of a new variable \( u \) and find the corresponding \( du \).
• Adjust the limits of integration to match the \( u \) transformation; for the example \( u = x^r \), each limit \( x = a \) and \( x = b \) becomes \( u = a^r \) and \( u = b^r \), respectively.
• Integrate the function from the new limits in terms of \( u \), and find the result.
• Transform the final result back to the original variable \( x \), if necessary, or apply the inverse directly with the new limits.

Using these steps helps ensure that integrals are evaluated correctly, keeping all transformations consistent with the integral's initial conditions.