Problem 37 - Calculus: One and Several Variables

Question

Calculate. $$\int \frac{1}{x} \arcsin (\ln x) d x$$

Step-by-Step Solution

Answer

\[\int \frac{1}{x} \arcsin (\ln x) dx = \arcsin (\ln x) \cdot \ln x - \sqrt{1 - (\ln x)^2} + C\]

Step 1: Identifying the parts

Let \( u = \arcsin (\ln x) \) and \( dv = \frac{1}{x} dx \). Then compute \( du \) and \( v \). Differentiating \( u \) with respect to \( x \) gives \( du = \frac{1}{x \sqrt{1-\ln^2 x}} dx \). Integrating \( dv \) gives \( v = \ln x \).

Step 2: Apply the integration by parts formula

Applying the integration by parts formula \(\int u dv = uv - \int v du \), we obtain: \(\int \frac{1}{x} \arcsin (\ln x) dx = u \cdot v - \int v * du = \arcsin (\ln x) \cdot \ln x - \int \frac{\ln x}{x \sqrt{1-\ln^2 x}} dx\).

Step 3: Further simplify the integral

The integral \(\int \frac{\ln x}{x \sqrt{1-\ln^2 x}} dx \) can be evaluated by substituting \( t = \ln x \), which simplifies it to \( \int \frac{t}{\sqrt{1 - t^2}} dt \). This is now a standard integral which evaluates to \( - \sqrt{1 - t^2} + C \), where \( C \) is the constant of integration.

Step 4: Substitute back the variables

Substitute back \( t = \ln x \) to obtain the final answer, \( - \sqrt{1 - (\ln x)^2} + C \).

Key Concepts

Definite Integrals

Definite integrals provide a way to calculate the area under a curve within a specific interval. Unlike indefinite integrals, which give a family of functions, definite integrals produce a numerical value. This value represents the net area between the curve and the x-axis.

When working with definite integrals, the boundaries of integration, denoted as limits, are crucial. They are placed on the integral sign from the lower limit to the upper limit. For example, to find the area under the function from \(a\) to \(b\), we write it as \(\int_a^b f(x) \,dx\).

Calculating definite integrals often involves the Fundamental Theorem of Calculus. This theorem links the process of differentiation and integration, allowing for the evaluation of an integral using an antiderivative.

• First, find an antiderivative of the function.
• Then, subtract the antiderivative evaluated at the lower limit from the antiderivative evaluated at the upper limit.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. The natural logarithm, represented as \ln x\, is a common example where the base is \e\. Logarithms help in dealing with exponential growth and decay, making calculations involving large numbers more manageable.

Key properties of logarithms include:

• Product rule: \ln(ab) = \ln a + \ln b\
• Quotient rule: \ln\left(\frac{a}{b}\right) = \ln a - \ln b\
• Power rule: \ln(a^b) = b\ln a\

When integrating functions involving \ln x\, we often use integration techniques such as substitution or integration by parts. In this exercise, \ln x\ was integrated to yield \ln x\, demonstrating its distinct behavior when compared to other functions.

The derivative of the natural logarithm is given by \(\frac{d}{dx}\ln x = \frac{1}{x}\). This fact is essential in calculus, especially when performing differentiation and integration involving logarithmic functions.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the roles of inputs and outputs for standard trigonometric functions. They are vital for solving equations involving trigonometric relationships or for integrating expressions that include trigonometric forms. For example, \arcsin x\ is the inverse of the sine function, giving the angle whose sine is \x\.

In inverse trigonometric functions, certain properties hold:

• Range is limited to principal values. For \arcsin x\, it lies between \[-\frac{\pi}{2}, \frac{\pi}{2}\]\.
• They have specific integration and differentiation formulas.

Their derivatives are often unique and essential for calculus applications. For example, the derivative of \arcsin x\ is \(\frac{d}{dx} \arcsin x = \frac{1}{\sqrt{1-x^2}}\).

Inverse trigonometric functions frequently appear in integration problems, especially involving substitution techniques where substituting a trigonometric identity simplifies the integral.