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Chapter 7: Problem 12

Perform the indicated integrations. \(\int_{2}^{3} \frac{d t}{t^{2} \sqrt{t^{2}-1}}\)

Short Answer

Expert verified
The integral evaluates to \(\frac{4\sqrt{2} - 3\sqrt{3}}{6}\).

Step by step solution

01

Identify Substitution for Integration

The integrand is in the form \[ \frac{1}{t^2 \sqrt{t^2 - 1}} \]This suggests that a trigonometric substitution might simplify the integral. For this form, let us use\[ t = \sec(\theta) \]which implies that \[ dt = \sec(\theta) \tan(\theta) d\theta. \]
02

Change Limits of Integration

Convert the limits of integration from terms of \(t\) to \(\theta\). When \(t = 2\), then \[ 2 = \sec(\theta) \Rightarrow \theta = \cos^{-1}\left(\frac{1}{2}\right). \]For \(t = 3\),\[ 3 = \sec(\theta) \Rightarrow \theta = \cos^{-1}\left(\frac{1}{3}\right). \]
03

Substitute and Simplify the Integral

By substitution, the integral becomes\[ \int_{\cos^{-1}\left(\frac{1}{2}\right)}^{\cos^{-1}\left(\frac{1}{3}\right)} \frac{\sec(\theta) \tan(\theta)}{\sec^2(\theta) \cdot \tan(\theta)} d\theta \]simplifying to\[ \int_{\cos^{-1}\left(\frac{1}{2}\right)}^{\cos^{-1}\left(\frac{1}{3}\right)} \frac{1}{\sec(\theta)} d\theta = \int_{\cos^{-1}\left(\frac{1}{2}\right)}^{\cos^{-1}\left(\frac{1}{3}\right)} \cos(\theta) d\theta. \]
04

Integrate and Evaluate

The integral of \(\cos(\theta)\) is \(\sin(\theta)\). Evaluating this from \(\cos^{-1}\left(\frac{1}{2}\right)\) to \(\cos^{-1}\left(\frac{1}{3}\right)\), we get\[ \sin\left(\cos^{-1}\left(\frac{1}{3}\right)\right) - \sin\left(\cos^{-1}\left(\frac{1}{2}\right)\right). \]Using the identity \( \sin(\cos^{-1}(x)) = \sqrt{1-x^2} \), compute:\[ \sin\left(\cos^{-1}\left(\frac{1}{3}\right)\right) = \sqrt{1-\left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}, \]\[ \sin\left(\cos^{-1}\left(\frac{1}{2}\right)\right) = \sqrt{1-\left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}. \]
05

Calculate the Result

Substitute back to find the final result:\[ \frac{2\sqrt{2}}{3} - \frac{\sqrt{3}}{2} = \frac{4\sqrt{2} - 3\sqrt{3}}{6}. \]Compute this expression to approximate the numerical value if necessary.

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Key Concepts

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

Definite Integration
Definite integration involves finding the integral of a function over a specific interval. This process calculates the net area under the curve of the function from one point to another. It's important to note that in definite integration:
  • The limits of integration are specified, typically as a lower and an upper bound.
  • The result is a number, unlike indefinite integration which results in a function.
  • One uses the Fundamental Theorem of Calculus to evaluate the integral at these bounds.
In the original exercise, we evaluate the definite integral of \(\int_{2}^{3} \frac{d t}{t^{2} \sqrt{t^{2}-1}}\)between the limits 2 and 3. The aim is to compute the total area under the curve \( \frac{1}{t^{2} \sqrt{t^{2}-1}} \)from \( t = 2 \)to \( t = 3. \)"Definite" means we're interested in this specific interval, and the integration will give us a concrete value.
Integration by Substitution
Integration by substitution is a crucial method used to simplify complex integrals by changing variables. This technique often transforms the integral into a more manageable form. Here are the steps involved:
  • Identify the substitution: This involves finding a trigonometric or algebraic function that can simplify the integral.
  • Change variables: Substitute the chosen function and its differential into the integral.
  • Back-substitute: Once integrated, revert to the original variable if necessary.
In our exercise, we used trigonometric substitution with \( t = \sec(\theta) \)because it directly addressed the \( \sqrt{t^2 - 1} \)term in the integrand. This simplification allowed us to rewrite the integral in terms of \( \theta \), facilitating easier evaluation. Moreover, the limits of integration were adjusted from \( t \) terms to \( \theta \) terms, ensuring correct bounds after substitution.
Integration Techniques
The process of integration often involves choosing from various techniques to make calculations manageable. Key techniques include:
  • Trigonometric Substitution: Used when encountering square roots of quadratic expressions (e.g., \( \sqrt{t^2 - 1} \)). This can simplify integrals into basic trigonometric forms.
  • Partial Fraction Decomposition: Useful for rational functions. It involves breaking down complex fractions into simpler parts.
  • Integration by Parts: Helpful for products of functions, using a formula akin to the product rule for differentiation.
For the given integral, trigonometric substitution was not just an option but a necessity. This method was ideal because the expression \( \sqrt{t^2 - 1} \)fits the perfect trigonometric identity \( \sec^2(\theta) - 1 = \tan^2(\theta) \).Identifying the right technique is essential for simplifying the integral and achieving an accurate result efficiently.

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Most popular questions from this chapter

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{(\sin t)\left(4 \cos ^{2} t-1\right)}{(\cos t)\left(1+2 \cos ^{2} t+\cos ^{4} t\right)} d t\)

Evaluate \(\int_{0}^{2 \pi} \frac{x|\sin x|}{1+\cos ^{2} x} d x .\) Hint: Make the substitution \(u=x-\pi\) in the definite integral and then use symmetry properties.

Use the method of partial fraction decomposition to perform the required integration. \(\int_{1}^{5} \frac{3 x+13}{x^{2}+4 x+3} d x\)

Find the length of the curve \(y=\ln (\cos x)\) between \(x=0\) and \(x=\pi / 4\).

For the differential equation \(\frac{d y}{d x}-\frac{y}{x}=x^{2}, x>0\), the integrating factor is \(e^{\int(-1 / x) d x} .\) The general antiderivative \(\int\left(-\frac{1}{x}\right) d x\) is equal to \(-\ln x+C .\) (a) Multiply both sides of the differential equation by \(\exp \left(\int\left(-\frac{1}{x}\right) d x\right)=\exp (-\ln x+C), \quad\) and show that \(\exp (-\ln x+C)\) is an integrating factor for every value of \(C .\) (b) Solve the resulting equation for \(y\), and show that the solution agrees with the solution obtained when we assumed that \(C=0\) in the integrating factor.
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