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Use an appropriate substitution and then a trigonometric substitution to evaluate the integrals. $$\int_{1}^{e} \frac{d y}{y \sqrt{1+(\ln y)^{2}}}$$

Short Answer

Expert verified
The integral evaluates to \( \ln(\sqrt{2} + 1) \).

Step by step solution

01

Choose the Substitution

We start by selecting a substitution that simplifies the integrand. Let \( x = \ln y \), so we have \( dx = \frac{1}{y} \, dy \). This substitution turns the integral bounds from \( y = 1 \) to \( y = e \) into \( x = \ln 1 = 0 \) and \( x = \ln e = 1 \). Substituting these into the integral yields:\[ \int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \, dx. \]
02

Apply Trigonometric Substitution

To further simplify \( \int \frac{1}{\sqrt{1+x^2}} \, dx \), we use the trigonometric substitution \( x = \tan(\theta) \), which implies \( dx = \sec^2(\theta) \, d\theta \) and \( \sqrt{1+x^2} = \sec(\theta) \). Substituting these in, the integral becomes:\[ \int \frac{\sec^2(\theta)}{\sec(\theta)} \, d\theta = \int \sec(\theta) \, d\theta. \]
03

Integral of the Trigonometric Function

The integral of \( \sec(\theta) \) is well-known and is given by:\[ \int \sec(\theta) \, d\theta = \ln |\sec\theta + \tan\theta| + C. \]
04

Convert Back to Original Variables

We need to convert \( \theta \) back to \( x \) and \( y \). Since \( x = \tan(\theta) \), we know \( \theta = \tan^{-1}(x) \) and \( \sec\theta = \sqrt{1+x^2} \). Therefore, the expression becomes:\[ \ln |\sqrt{1+x^2} + x|. \]
05

Evaluate the Integral with New Limits

Use the substitution limits, which are from \( x = 0 \) to \( x = 1 \). Therefore, we evaluate\[ \ln |\sqrt{1+x^2} + x| \bigg|_{0}^{1}. \]Plug in the limits:\[ \ln |\sqrt{2} + 1| - \ln |1|. \]
06

Final Step: Simplify the Result

Simplifying the expression, we find the evaluation:\[ \ln(\sqrt{2} + 1) - 0 = \ln(\sqrt{2} + 1). \]Thus, the solution evaluates to \( \ln(\sqrt{2} + 1) \).

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Key Concepts

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

Trigonometric Substitution
Trigonometric substitution is a powerful technique used in calculus to solve integrals involving square roots, especially those with expressions such as \(\sqrt{a^2 + x^2}\), \(\sqrt{a^2 - x^2}\), or \(\sqrt{x^2 - a^2}\). This method takes advantage of the Pythagorean identities and trigonometric functions to simplify the integration process.

In our case, the integral \(\int \frac{1}{\sqrt{1+x^2}} \, dx\) suggests the use of tangent substitution. Here, we set \(x = \tan(\theta)\), which converts the expression to involve \(\sec(\theta)\), since \(\sqrt{1 + x^2} = \sec(\theta)\). This substitution simplifies the integral and transforms it into a more manageable form.

The matching identities required are:
  • \(x = \tan(\theta)\)
  • \(dx = \sec^2(\theta) \, d\theta\)
  • \(\sqrt{1+x^2} = \sec(\theta)\)
These adjustments pave the way for the calculation of the integral by converting the complicated expression into simpler trigonometric functions, hence making the problem more solvable.
Definite Integrals
Definite integrals are a fundamental concept in calculus that calculate the net area under a curve within specified limits, providing a number rather than an expression. They extend the concept of antiderivatives to determine the actual value of a function across a certain interval.

In this exercise, we worked with a definite integral, \(\int_{1}^{e} \frac{dy}{y \sqrt{1+ (\ln y)^{2}}}\), which required changing the variables using substitution. Initially, we replaced \(y\) with \(\ln y\), transforming the initial bounds \(y=1\) and \(y=e\) into \(x=0\) and \(x=1\), respectively. This step is crucial as it sets the scene for integrating over the newly substituted variable.

After applying trigonometric substitution, we resolved the new integral \(\int_{0}^{1} \frac{1}{\sqrt{1+x^2}} \, dx\) and evaluated it between the bounds of 0 and 1. Calculating this definite integral gave us the final answer \(\ln(\sqrt{2} + 1)\). This process showcases the practical use of definite integrals in determining the precise value of original expressions within given limits.
Integration Techniques
Integration techniques are various methods in calculus that aid in solving integrals which are not straightforward. Each technique has its strategy and is applicable depending on the form of the function to be integrated.

For the given problem, we combined multiple methods to tackle the complex integral effectively.
  • **Substitution Method:** Initially used to align the integral closer to a form manageable by trigonometric substitution. Here, \(x = \ln y\) was selected, simplifying our given expression substantially.
  • **Trigonometric Substitution:** Employed to manage the square root within the integral confidently by transforming \(x = \tan(\theta)\), this took advantage of trigonometric identities like \(\sec^2(\theta)\) simplifying it to a standard integral of \(\sec(\theta)\).
By strategically combining these methods, the integration became significantly more feasible. Understanding when and how to apply these techniques is key in mastering calculus integration.

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