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Evaluate the integrals in Exercises \(1-14\) $$ \int \frac{d x}{\sqrt{4 x^{2}-49}}, \quad x>\frac{7}{2} $$

Short Answer

Expert verified
The integral evaluates to \( \frac{7}{2} \ln \left| \frac{2x + \sqrt{4x^2 - 49}}{7} \right| + C.\)

Step by step solution

01

Recognize the Integral Form

Observe that the given integral is of the form \( \int \frac{dx}{\sqrt{a^2 - x^2}} \), which suggests it relates to the standard inverse trigonometric functions. However, it needs to be manipulated to fit this form.
02

Simplify the Expression Inside the Square Root

Rewrite the expression inside the square root: \( 4x^2 - 49 \) as \( (2x)^2 - 7^2 \) to match the form \( u^2 - a^2 \). This suggests a trigonometric substitution might be appropriate for simplification.
03

Choose an Appropriate Substitution

Set \( 2x = 7 \sec(\theta) \), as \( \sec(\theta) \) is appropriate for expressions of the form \( \sec^2(\theta) - 1 = \tan^2(\theta) \). This implies that \( dx = \frac{7}{2} \sec(\theta) \tan(\theta) \, d\theta \).
04

Substitute in the Integral

Substitute \( 2x = 7\sec(\theta) \) and \( dx = \frac{7}{2}\sec(\theta)\tan(\theta)\,d\theta \) into the integral:\[\int \frac{dx}{\sqrt{4x^2 - 49}} = \int \frac{\frac{7}{2}\sec(\theta)\tan(\theta)\,d\theta}{\sqrt{(2x)^2 - 7^2}} = \int \frac{\frac{7}{2}\sec(\theta)\tan(\theta)\,d\theta}{7\tan(\theta)}\]
05

Simplify the Integral

The integral simplifies to:\[ \int \frac{7}{2}\sec(\theta) \, d\theta \]Upon canceling \( \tan(\theta) \).
06

Evaluate the Integral

Evaluate the simplified integral: \[\int \frac{7}{2}\sec(\theta)\, d\theta = \frac{7}{2} \ln |\sec(\theta) + \tan(\theta)| + C\]where \( C \) is the constant of integration.
07

Back-substitute to Original Variables

Recall that \( 2x = 7\sec(\theta) \), so \( \sec(\theta) = \frac{2x}{7} \). Also, \( \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{2x}{7}\right)^2 - 1} \). Substitute these back to get:\[\frac{7}{2} \ln \left| \frac{2x}{7} + \sqrt{\left(\frac{2x}{7}\right)^2 - 1} \right| + C\]
08

Simplify the Expression

Simplify the back-substituted form by entering the expressions of absolute values:\[\frac{7}{2} \ln \left| \frac{2x}{7} + \frac{\sqrt{4x^2 - 49}}{7} \right| + C\]This is the final result of the original integral in terms of \( x \).

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions, such as arc sine (\(\sin^{-1}\)), arc cosine (\(\cos^{-1}\)), and arc tangent (\(\tan^{-1}\)), are essential tools in calculus for finding angles that correspond to given trigonometric ratios. In certain integral problems, these functions appear when we use trigonometric substitution to simplify integrals.For instance, consider the integral \(\int \frac{dx}{\sqrt{a^2 - x^2}}\). This is directly related to the arc sine function because the derivative of \(\sin^{-1}(x)\) is \(\frac{1}{\sqrt{1-x^2}}\). Therefore, a substitution that simplifies the integral into this form often leads to inverse trigonometric solutions.In our specific problem, identifying and aiming for this structure was key. By rewriting the expression under the square root, we manipulated the problem into a form that could be tackled using inverse trigonometric functions, ultimately leading to a simpler evaluation.
Integral Calculus
Integral calculus deals with the concept of integration, which is essentially the reverse process of differentiation. It allows us to find accumulated quantities, areas under curves, and solutions to differential equations.In particular, indefinite integrals (also known as antiderivatives) don't include limits and result in a family of functions plus an arbitrary constant \(C\). Definite integrals, on the other hand, are evaluated between two bounds and yield a numerical result that represents the accumulated sum over an interval.Understanding integral calculus is crucial for dealing with expressions involving trigonometric functions. The exercise presented demonstrates how one can manipulate an expression to find its integral using substitution - a common technique used to simplify the integration process and transform complex problems into recognizable forms.
Definite Integrals
Definite integrals compute the net area between the graph of a function and the x-axis, over a specific interval \([a, b]\). They are crucial for solving real-world problems involving accumulations like distance, area, and work.The integral \(\int_{a}^{b} f(x) \, dx\) involves calculating the limit of sums of areas of rectangles, thus providing a precise numeric value unlike indefinite integrals which include an arbitrary constant. The exercise we discussed focuses on indefinite integration, but possessing a grasp of definite integrals sharpens one's understanding of how the accumulated change over an interval works and aids in comprehending the real-world application of integral calculus.
Trigonometric Identities
Trigonometric identities are equations relating to the trigonometric functions that are true for any angle. They are a fundamental aspect of trigonometry that simplify complex expressions and solve trigonometric equations.In integral calculus, these identities can transform non-trivial expressions into more manageable forms. For instance, the identity \(\sec^2(\theta) - 1 = \tan^2(\theta)\) is utilized during substitutions in integration problems. This transformation is crucial when we used \(2x = 7 \sec(\theta)\) in the exercise.Trigonometric identities are also used to backtrack from a substituted variable to the original terms of the integral after evaluation. Mastering these identities simplifies solving trigonometric substitution problems and enhances one's problem-solving toolkit in calculus.

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Most popular questions from this chapter

\begin{equation} \begin{array}{l}{\text { Consider the infinite region in the first quadrant bounded by the }} \\ {\text { graphs of } y=\frac{1}{x^{2}}, y=0, \text { and } x=1 .} \\ {\text { a. Find the area of the region. }} \\ {\text { b. Find the volume of the solid formed by revolving the region }} \\ {\text { (i) about the } x \text { -axis; (ii) about the } y \text { -axis. }}\end{array} \end{equation}

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\begin{equation} \begin{array}{l}{\text { Consider the infinite region in the first quadrant bounded by the }} \\ {\text { graphs of } y=\frac{1}{\sqrt{x}}, y=0, x=0, \text { and } x=1} \\ {\text { a. Find the area of the region. }} \\ {\text { b. Find the volume of the solid formed by revolving the region }} \\ {\text { (i) about the } x \text { -axis; (ii) about the } y \text { -axis. }}\end{array} \end{equation}

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