Question

In the following exercises, integrate using the indicated substitution. $$ \int \ln (x) \frac{\sqrt{1-(\ln x)^{2}}}{x} d x ; u=\ln x $$

Short answer

The integral evaluates to \(-\frac{1}{3} (1-(\ln x)^2)^{3/2} + C\).

Step-by-step solution

Identify the Substitution

The substitution given in the problem is \( u = \ln x \). This implies that the next step is to find \( du \), and since \( du/dx = 1/x \), we have \( du = \frac{1}{x} dx \).

Substitute Variables

After substituting \( u = \ln x \) and \( du = \frac{1}{x} dx \) into the integral, we rewrite the integral as follows:\[\int \ln(x) \frac{\sqrt{1-(\ln x)^{2}}}{x} dx = \int u \sqrt{1-u^2} \, du\]

Integrate Using Substitution

Now, we focus on integrating \( \int u \sqrt{1-u^2} \, du \). To do this, use the substitution \( v = 1-u^2 \), then \( dv = -2u \, du \). Thus, \( u \, du = -\frac{1}{2} dv \). Substitute into the integral:\[\int u \sqrt{1-u^2} \, du = -\frac{1}{2} \int \sqrt{v} \, dv\]

Solve the New Integral

Now, integrate \( -\frac{1}{2} \int \sqrt{v} \, dv \):\[-\frac{1}{2} \int v^{1/2} \, dv = -\frac{1}{2} \times \frac{2}{3} v^{3/2} + C = -\frac{1}{3} v^{3/2} + C\]

Back-substitute the Original Variable

Replace \( v \) with \( 1-u^2 \) and \( u \) with \( \ln x \) to return to the original variable:\[-\frac{1}{3} (1-u^2)^{3/2} + C = -\frac{1}{3} (1-(\ln x)^2)^{3/2} + C\]

Final Answer

The integrated form is:\[\int \ln(x) \frac{\sqrt{1-(\ln x)^{2}}}{x} dx = -\frac{1}{3} (1-(\ln x)^2)^{3/2} + C\]This is the final answer after performing the integration and substituting back the original variables.

Key concepts

Substitution Method

The substitution method is a powerful technique in calculus, especially useful for solving complex integrals. Often, integrals become simpler when rewritten in terms of a new variable. This new variable acts as a substitute for part of the original function, making it easier to handle.
One common scenario for using substitution is when you have a nested or composite function to integrate, which involves another function. Essentially, you look for a part of the original integral that, when differentiated, appears elsewhere in the integral. This part is set equal to a new variable, often denoted by \( u \).
Once you've defined \( u \), the next step is finding \( du \), the derivative of \( u \), in terms of the original variable. This involves expressing \( dx \) (the differential in the integral) in terms of \( du \).
When using substitution, remember:

• Identify the substitution: Choose what part of the integral to replace.
• Rewrite the integral: Express everything in terms of the new variable.
• Change the differential: Replace \( dx \) with \( du \) based on your substitution.
• Integrate in the new variable: Solve the simpler integral in terms of \( u \).
• Return to the original variable: Substitute back to express the result in terms of the original variable.

Using this method helps transform complex integrals into more manageable forms.

Integration by Substitution

Integration by substitution is a dedicated approach to tackle integrals involving intricate functions. By its nature, substitution aims to make the integral look simpler and thus easier to integrate. The idea is to change variables in a strategic way so the integral resembles a more straightforward form.
Take our exercise as an example: the integral involves \( \ln(x) \), which is efficiently handled by letting \( u = \ln(x) \). This substitution transforms the original integral to a simpler form: \( \int u \sqrt{1-u^2} \, du \).
In the next step, another substitution \( v = 1-u^2 \) helps further simplify it so you can integrate without additional complexity. Substitution requires:

• Recognizing the right substitution to make the integral simpler.
• Performing differentiation and adjustments to accommodate the substitution.
• Solving the transformed integral before reverting to the original variable.

Remember, integration by substitution is analogous to reversing the chain rule in differentiation.

Definite and Indefinite Integrals

Integrals come in two flavors: definite and indefinite. Both play a crucial role in calculus, with slight but important distinctions.
Indefinite integrals refer to finding the antiderivative of a function. It doesn't have limits of integration, and you'll always include a constant of integration, \( C \), since the antiderivative is only defined up to a constant. This constant represents any function that differs only by a constant from the indefinite integral.
In our example, the final indefinite integral was expressed with \( C \) to account for all these possibilities. It's typically written as \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative of \( f(x) \).
On the other hand, definite integrals compute the net area under a curve within given boundaries or limits. Here, instead of a constant, the evaluation involves calculating the difference of the antiderivative at two points. The notation is \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
Concisely, while indefinite integrals give a family of functions, definite integrals provide numerical results. However, regardless of the type, the fundamental techniques used, such as substitution, remain essential across both.