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Chapter 7: Problem 42

First make a substitution and then use integration by parts to evaluate the integral. \( \displaystyle \int \frac{\arcsin (\ln x)}{x} dx \)

Short Answer

Expert verified
The integral evaluates to \( \ln x \cdot \arcsin(\ln x) + \sqrt{1-(\ln x)^2} + C \).

Step by step solution

01

Choose a Substitution

Start with the substitution \( u = \ln x \). This means that \( du = \frac{1}{x} dx \), simplifying the expression \( \int \frac{\arcsin(\ln x)}{x} dx \) to \( \int \arcsin(u) \, du \).
02

Apply Integration by Parts

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Choose \( v = \arcsin(u) \) and \( dw = du \). Therefore, \( dv = 1 \cdot du \) and integrate to find \( w = u \). Now, apply the integration by parts: \[ \int \arcsin(u) \, du = u \cdot \arcsin(u) - \int u \cdot \frac{1}{\sqrt{1-u^2}} \, du \].
03

Solve the Second Integral

The remaining integral is \( \int \frac{u}{\sqrt{1-u^2}} \, du \). By substitution, let \( v = 1-u^2 \), then \( dv = -2u \, du \). So, \( u \, du = -\frac{1}{2} \, dv \). Therefore, the integral becomes \(-\frac{1}{2} \int \frac{1}{\sqrt{v}} \, dv \), which simplifies to \(-\sqrt{v} = -\sqrt{1-u^2} \).
04

Combine Results

Combining all terms, we find: \[ \int \arcsin(u) \, du = u \cdot \arcsin(u) + \sqrt{1-u^2} + C \]. Substitute back \( u = \ln x \): \[ \ln x \cdot \arcsin(\ln x) + \sqrt{1-(\ln x)^2} + C \].
05

Final Expression

Thus, the integral \( \int \frac{\arcsin(\ln x)}{x} dx \) evaluates to \( \ln x \cdot \arcsin(\ln x) + \sqrt{1-(\ln x)^2} + C \), where \( C \) is the integration constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Substitution Method
The substitution method is a powerful technique used to simplify complex integrals by transforming them into a more familiar form. It involves substituting a part of the integral with a new variable, which reduces the complexity.
  • Choose an appropriate substitution: Start by identifying a part of the integral that can be replaced with a single variable, making the overall expression simpler.
  • Relate the variables: Establish a relation between the new and original variables, such as expressing everything in terms of the new variable.
In the exercise, we used the substitution \( u = \ln x \), simplifying the integral to \( \int \arcsin(u) \ du \). Differentiating gives \( du = \frac{1}{x} \, dx \), making the expression much easier to handle.This transformation often paves the way for using other integration techniques, like integration by parts, to solve the integral.
Definite Integral
A definite integral is the evaluation of the integral of a function within specific limits. It represents the area under the curve of a function within those bounds.
  • Represented as \( \int_{a}^{b} f(x) \, dx \): The endpoints \( a \) and \( b \) are the limits of integration.
  • Calculates net area: Positive area contributes positively, and negative area contributes negatively, depending on the part of the x-axis it lies.
Although the provided exercise evaluates an indefinite integral (without limits), understanding definite integrals is essential, as they are widely used in applications like finding total distance traveled or the work done by a force.
Indefinite Integral
Indefinite integrals, also known as antiderivatives, represent a family of functions and bring about an arbitrary constant. These are not evaluated over a range but rather provide a general form for all possible values of the integral.
  • No bounds: Unlike definite integrals, they lack upper and lower limits, which means the result includes a constant \( C \) because the derivative of a constant is zero.
  • The general solution: The solution of an indefinite integral gives a general solution to differential equations.
In the exercise, the indefinite integral \( \int \frac{\arcsin(\ln x)}{x} \, dx \) was evaluated to yield a function plus a constant \( C \), accounting for all possible antiderivatives.
Arcsine Function
The arcsine function, denoted as \( \arcsin(x) \), is the inverse of the sine function and plays a crucial role in trigonometry and calculus.
  • Range and domain: The arcsine function returns angles, with outputs restricted between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), while taking inputs within \(-1 \leq x \leq 1\).
  • Usage: Commonly appears when dealing with integrals and in solving trigonometric equations that involve inverse trigonometric functions.
In this exercise, arcsine was utilized in the integral as part of the expression \( \int \arcsin(\ln x) \, du \). Rigorous handling of such inverse functions is essential, especially when paired with other calculus techniques like integration by parts.

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Most popular questions from this chapter

Use (a) the Trapezoidal Rule, (b) the Midpoint Rule, and (c) Simpson's Rule to approximate the given integral with the specified value of \( n \). (Round your answers to six decimal places.) \( \displaystyle \int_1^3 e^{\frac{1}{x}}\ dx \) , \( n = 8 \)

Explain why each of the following integrals is improper. (a) \( \displaystyle \int_1^2 \frac{x}{x - 1}\ dx \) (b) \( \displaystyle \int_0^\infty \frac{1}{1 + x^3}\ dx \) (c) \( \displaystyle \int_{-\infty}^\infty x^2 e^{-x^2}\ dx \) (d) \( \displaystyle \int_0^{\frac{\pi}{4}} \cot x\ dx \)

Find the values of \( p \) for which the integral converges and evaluate the integral for those values of \( p \). \( \displaystyle \int_0^1 \frac{1}{x^p}\ dx \)

Draw the graph of \( f(x) = \sin \left (\frac{1}{2} x^2 \right) \) in the viewing rectangle \( [0, 1] \) by \( [0, 0.5] \) and let \( I = \displaystyle \int_0^1 f(x)\ dx \). (a) Use the graph to decide whether \( L_2 \), \( R_2 \), \( M_2 \), and \( T_2 \) underestimate or overestimate \( I \). (b) For any value of \( n \), list the numbers \( L_n \), \( R_n \), \( M_n \), \( T_n \), and \( I \) in increasing order. (c) Compute \( L_5 \), \( R_5 \), \( M_5 \), and \( T_5 \). From the graph, which do you think gives the best estimate of \( I \)?

Sketch the graph of a continuous function on \( [0, 2] \) for which the Trapezoidal Rule with \( n = 2 \) is more accurate than the Midpoint Rule.
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