Question

Integral family 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 \(0< x <1\).

Short answer

Question: Find the integral of \(\int \frac{dx}{x\sqrt{1-x^{2r}}}\) using the substitution \(u = x^r\). Answer: \(\int \frac{dx}{x\sqrt{1-x^{2r}}} = -\frac{1}{r}\operatorname{sech}^{-1}(x^{r}) + C\)

Step-by-step solution

Apply the substitution \(u = x^r\)

Make the substitution \(u = x^r\). We then calculate the derivative of this substitution: $$\frac{du}{dx} = r x^{r-1}.$$ Now let's rewrite the original integral in terms of \(u\): $$\int \frac{dx}{x\sqrt{1-x^{2r}}} = \int \frac{du / (r x^{r-1})}{x\sqrt{1-u^2}}.$$

Simplify the integral in terms of \(u\)

Now, we can simplify the integral expression by canceling out the terms \(x^{r-1}\) in the numerator and denominator: $$\int \frac{du / (r x^{r-1})}{x\sqrt{1-u^2}} = \int \frac{du}{r\sqrt{1-u^2}}.$$

Integrate with respect to \(u\)

Now we can perform the integration: $$\int \frac{du}{r\sqrt{1-u^2}} = -\frac{1}{r} \int \frac{du}{\sqrt{1-u^2}}.$$ The integral on the right-hand side is a standard integral: $$\int \frac{du}{\sqrt{1-u^2}} = \operatorname{sinh}^{-1}(u) + C_1.$$ So, now we have: $$-\frac{1}{r} \int \frac{du}{\sqrt{1-u^2}} = -\frac{1}{r}\operatorname{sinh}^{-1}(u) + \frac{C_1}{r}.$$

Substitute \(x^r\) back in for \(u\)

In the final step, we substitute \(x^r\) back in for \(u\): $$-\frac{1}{r}\operatorname{sinh}^{-1}(x^{r}) + \frac{C_1}{r} = -\frac{1}{r}\operatorname{sech}^{-1}(x^{r}) +C.$$ Here, we have used the fact that \(\operatorname{sinh}^{-1}(x) = \operatorname{sech}^{-1}(x)\) and \(C\) can represent the constant of integration. Thus, the final solution is: $$\int \frac{dx}{x\sqrt{1-x^{2r}}} = -\frac{1}{r}\operatorname{sech}^{-1}(x^{r}) + C.$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used for solving integrals involving square roots or quadratic expressions. It's extremely useful when dealing with integrals that contain expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \). This method aims to simplify these expressions by substituting a trigonometric identity to convert the integral into a more easily solvable form.

Here are some typical substitutions you might consider:

• For \( \sqrt{1 - x^2} \), use the substitution \( x = \sin{\theta} \), where \( \theta \) is the angle such that the identity holds true.
• For \( \sqrt{x^2 + a^2} \), use the substitution \( x = a \tan{\theta} \).
• For \( \sqrt{x^2 - a^2} \), consider \( x = a \sec{\theta} \).

Once the substitution is made, the new expression will often result in a standard trigonometric integral, which is easier to tackle. Reverting back to the variable \( x \) is a crucial final step to ensure that the solution is expressed in terms of the original variable.

In our original exercise, after the substitution \( u = x^r \) is made, the integral simplifies into a form that can be rewritten using inverse hyperbolic functions, showing the flexibility and usefulness of trigonometric substitution.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are the hyperbolic equivalents of inverse trigonometric functions. They are instrumental in integration, particularly when dealing with expressions involving \( \sqrt{1-u^2} \), similar to the inverse trigonometric function arcsin. The most commonly used inverse hyperbolic functions include \( \operatorname{sinh}^{-1}(x) \) and \( \operatorname{sech}^{-1}(x) \).

Inverse hyperbolic functions have distinct properties and are defined by unique logarithmic functions. For example, the inverse hyperbolic sine \( \operatorname{sinh}^{-1}(x) \) is defined as \( \ln(x + \sqrt{x^2 + 1}) \). These functions are particularly useful in integrals where hyperbolic identities simplify the expression

In our original problem, we encountered the integral \( \int \frac{du}{\sqrt{1-u^2}} \), which is a standard inverse hyperbolic function. Integrating this results in \( \operatorname{sinh}^{-1}(u) \). Upon substituting back \( x^r \) for \( u \), we utilized the transformation from \( \operatorname{sinh}^{-1} \) to \( \operatorname{sech}^{-1} \), demonstrating the interchangeability and application of inverse hyperbolic functions in integration.

Definite and Indefinite Integrals

Integrals are fundamental in calculus and come in two varieties: definite and indefinite.

Indefinite integrals, often represented as \( \int f(x) \, dx \), produce a function \( F(x) \) whose derivative is \( f(x) \). Indefinite integrals include a constant \( C \), representing an infinite number of potential solutions, as they don't have specific boundaries.

Definite integrals, denoted with limits, \( \int_{a}^{b} f(x) \, dx \), result in a specific numerical value. They calculate the accumulated sum or area under the curve \( f(x) \) between \( x = a \) and \( x = b \).

In our exercise, we dealt with an indefinite integral. We ended with a general solution, including the constant of integration \( C \), as it was evaluated without specific limits. This highlights the indefinite integral's role in finding antiderivatives or "families" of functions, essential for understanding dynamic systems or changes in processes over intervals.