Question

Evaluate the following integrals: $$ \int \frac{\ln x d x}{x \sqrt{1-4 \ln x-\ln ^{2} x}} $$

Short answer

Question: Evaluate the integral: $$ \int \frac{\ln x dx}{x \sqrt{1-4 \ln x-\ln ^{2} x}} $$ Answer: $$ -2\cos\left(\arcsin\left(\ln x + \frac{1}{2}\right)\right) -\arcsin\left(\ln x + \frac{1}{2}\right) + C $$

Step-by-step solution

Substitution

Let \(u = 2\ln x + 1\). We will find \(du\) and make the substitution. First, differentiate \(u\) with respect to \(x\): $$ \frac{du}{dx} = 2\frac{1}{x}\implies du = \frac{2 dx}{x} $$ Now solve for \(x\) in terms of \(u\): $$u-1=2\ln x\implies x=e^{(u-1)/2}$$ Now, make the substitution in the integral: $$ \int \frac{\ln x d x}{x \sqrt{1-4 \ln x-\ln ^{2} x}} = \int \frac{(u-1)/2\cdot \frac{2du}{e^{(u-1)/2}}}{e^{(u-1)/2} \sqrt{1-(2u-(u-1))^2}} $$ The \(e^{(u-1)/2}\) term cancels out, so the new integral to solve is: $$ \int \frac{(u-1) du}{\sqrt{4 - u^2}} $$

Solve the Integral in terms of \(u\)

To solve the integral: $$ \int \frac{(u-1) du}{\sqrt{4-u^2}} $$ We will use the trigonometric substitution \(u=2\sin\theta\). We then have \(du=2\cos\theta d\theta\). Substituting, we get: $$ \int \frac{(2\sin\theta-1)(2\cos\theta d\theta)}{2\cos\theta} = 2\int (\sin\theta-0.5) d\theta $$ Now, we can integrate with respect to \(\theta\): $$ 2\int (\sin\theta-0.5) d\theta = -2\cos\theta -\theta + C $$ Now we need to switch back to the \(u\) variable and then our original \(x\) variable.

Convert Back to Original Variable

First, recall that \(u=2\sin\theta\), so \(\sin\theta=\frac{u}{2}\). Now we find_theta in terms of \(u\): $$ \theta = \arcsin\left(\frac{u}{2}\right). $$ Substituting back into our equation, we get: $$ -2\cos\left(\arcsin\left(\frac{u}{2}\right)\right) -\arcsin\left(\frac{u}{2}\right) + C $$ Now recall that \(u=2\ln x + 1\), so we get: $$ -2\cos\left(\arcsin\left(\frac{2\ln x}{2} + \frac{1}{2}\right)\right) -\arcsin\left(\frac{2\ln x}{2} + \frac{1}{2}\right) + C $$ This gives us our final answer for the original integral: $$ -2\cos\left(\arcsin\left(\ln x + \frac{1}{2}\right)\right) -\arcsin\left(\ln x + \frac{1}{2}\right) + C $$

Key concepts

Integration Techniques

Integration is a fundamental concept in calculus that involves finding the antiderivative or the integral of a function. This allows us to find areas, volumes, central points, and many useful things, but it can sometimes be tricky to carry out. There are several integration techniques designed to make this process easier, especially when dealing with complex functions.

Common Techniques

• Substitution: Useful for functions that are the product of a function and its derivative, or when a change of variables simplifies the function.
• Integration by Parts: Useful for products of functions. It is based on the reverse of the product rule from differentiation.
• Partial Fractions: Breaking down fractions into simpler parts to integrate.
• Trigonometric Substitution: Replacing variables with trigonometric functions to simplify integrals involving square roots.
• Numerical Integration: Approximating the integral using methods like Simpson's rule when an integral cannot be solved analytically.

Choosing the right technique is crucial and can greatly simplify the calculation of an integral. The exercise provided uses substitution to simplify the integral before employing trigonometric substitution, showcasing how multiple techniques can be needed for a single problem.

Trigonometric Substitution

Trigonometric substitution is a technique often used when an integral involves a radical expression or when the integrand suggests a trigonometric identity. The idea is to substitute a trigonometric function for a variable to leverage the properties of trigonometric functions that can simplify the integral.

How It Works

For example, if an integral contains a \( \sqrt{a^2 - x^2} \) term, one might use the substitution \( x = a\sin(\theta) \), because \( \sin^2(\theta) \) will conveniently convert \( a^2 - (a\sin(\theta))^2 \) to \( a^2\cos^2(\theta) \), which is easier to integrate. Upon solving the integral in terms of \( \theta \) or the trigonometric function, the last step is to convert back to the original variable, often involving arcsine, arccosine, or arctangent functions. This step is critical and is well demonstrated in the provided exercise where after applying trigonometric substitution, the solution is converted back from \( \theta \) to \( u \) and ultimately to \( x \) to express the integral in terms of the original variable.

Logarithmic Integration

Logarithmic integration is a technique specifically used to integrate functions involving logarithms. This can either involve integrating the logarithmic function directly or dealing with integrands where the logarithm is part of a larger function.

Integrating a Simple Logarithm

For the integral of \( \ln(x) \), the integration by parts method is typically used. This is because \( \ln(x) \) does not have an obvious antiderivative.

Complex Logarithmic Functions

In more complex scenarios, like the one in the exercise, we often perform a substitution to simplify the function into a form that can then be integrated directly or that allows for additional techniques like trigonometric substitution. The given exercise required substituting to simplify the integrand before proceeding with the integration. The substitution not only reduced the complexity of the integral but also enabled the use of trigonometric substitution to solve it, highlighting the versatility of integration techniques in calculus.