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Chapter 7: Problem 37

Evaluate the following integrals. $$\int \frac{\sqrt{9 x^{2}-25}}{x^{3}} d x, x>\frac{5}{3}$$

Short Answer

Expert verified
Question: Evaluate the integral: $$ \int \frac{\sqrt{9x^2-25}}{x^3} dx. $$ Solution: The evaluated integral is given by: $$ \frac{4}{45}\left(5\sec^{-1}\left(\frac{\sqrt{2} x}{5}\right)-\sqrt{2}\right)+C. $$

Step by step solution

01

Trigonometric Substitution

Since the integrand involves the square root of a quadratic expression, we use the trigonometric substitution: $$ x=\frac{5}{\sqrt{2}}\sec\theta. $$ Now, differentiate x with respect to 胃 to prepare for substitution: $$ \frac{dx}{d\theta}=\frac{5}{\sqrt{2}}\sec\theta\tan\theta. $$
02

Substitution of x and dx

Now, we substitute the expressions for x and dx in terms of 胃 into the integral: $$ \begin{aligned} \int \frac{\sqrt{9x^2-25}}{x^3} dx &= \int \frac{\sqrt{9\left(\frac{5}{\sqrt{2}} \sec\theta\right)^2-25}}{\left(\frac{5}{\sqrt{2}}\sec\theta\right)^3} \left(\frac{5}{\sqrt{2}}\sec\theta\tan\theta\right) d\theta \\ &= \int \frac{\sqrt{9\left(\frac{25}{2} \sec^2\theta\right)-25}}{\frac{125}{4\sqrt{2}}\sec^3\theta} \frac{5\sec\theta\tan\theta}{\sqrt{2}} d\theta. \end{aligned} $$
03

Simplify the integrand

Now, simplify the expression under the square root and cancel terms in the integrand: $$ \begin{aligned} &= \int \frac{\sqrt{\frac{225}{2} \sec^2\theta-25}}{\frac{125}{4\sqrt{2}}\sec^3\theta} \frac{5\sec\theta\tan\theta}{\sqrt{2}} d\theta \\ &= \int \frac{\sqrt{\frac{25(9\sec^2\theta-2)}}{2}}}{\frac{125}{4\sqrt{2}}\sec^3\theta} \frac{5\sec\theta\tan\theta}{\sqrt{2}} d\theta \\ &= \frac{20}{125\sqrt{2}} \int \frac{\sqrt{9\sec^2\theta-2}}{\sec^2\theta} \tan\theta d\theta. \end{aligned} $$
04

Integrate

Now, integrate the resulting expression: $$ \begin{aligned} &= \frac{20}{125\sqrt{2}} \int \frac{\sqrt{9\sec^2\theta-2}}{\sec^2\theta} \tan\theta d\theta \\ &= \frac{20}{125\sqrt{2}}\left[-\frac{2\sqrt{2}+5\theta}{9}\right] + C. \end{aligned} $$
05

Substitute back x

Finally, substitute back the original variable x using the trigonometric substitution: $$ \theta = \sec^{-1}\left(\frac{\sqrt{2} x}{5}\right). $$ Thus, $$ \begin{aligned} \int \frac{\sqrt{9x^2-25}}{x^3} dx &= \frac{20}{125\sqrt{2}}\left[-\frac{2\sqrt{2}+5\sec^{-1}\left(\frac{\sqrt{2} x}{5}\right)}{9}\right] + C \\ &= \frac{4}{45}(5\sec^{-1}\left(\frac{\sqrt{2} x}{5}\right)-\sqrt{2})+C. \end{aligned} $$

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Key Concepts

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

Integral Calculus
Integral calculus is a fundamental branch of calculus concerned with the concept of accumulation and area under curves. Unlike its cousin, differential calculus, which focuses on changes and slopes, integral calculus helps us understand total quantities. The core idea is to find the antiderivative, or integral, of a function, effectively reversing the process of differentiation.

In particular, when faced with an integral involving a complex expression, such as those involving radicals or trigonometric functions, special techniques can be employed. Trigonometric substitution is one such technique used for integrals that involve square roots of quadratic expressions.

For the example given:
  • The expression inside the square root, \(9x^2-25\), is re-framed using a trigonometric identity by substituting \(x\) with \(\frac{5}{\sqrt{2}}\sec\theta\). This simplifies the complexity involved in the square root.
  • This substitution takes advantage of the identity \(\sec^2\theta - 1 = \tan^2\theta\), which helps simplify the integration process.
Definite Integrals
Definite integrals are a specific type of integral that calculates the accumulated quantity bounded between two points. In contrast to indefinite integrals, which include a constant \(C\) due to their general form, definite integrals provide a specific numerical result. This is achieved by evaluating the antiderivative at the upper limit and subtracting the evaluation at the lower limit.

The integral from the given problem, once evaluated and simplified, demonstrates the concept of limits. Although the particular exercise focuses on an indefinite form for demonstration, applying definite integrals means calculating specific values where limits are defined. When facing definite integrals, each involves:
  • The function to be integrated, often requiring simplification or substitution for ease.
  • Limits of integration, indicating the specific section of the curve for accumulation.
Integral Simplification
Integral simplification is crucial in solving integrals, especially those involving complex functions or expressions. By using algebraic manipulation and calculus techniques, integrals can be expressed in a simpler, more manageable form before performing the actual integration.

In the context of the exercise, simplification occurs through sequencing the work:
  • Using trigonometric substitution to eliminate radicals.
  • Simplifying the resulting trigonometric expressions by canceling terms.
  • Breaking down the integrand to a point where the integration can be straightforwardly performed.
This process helps reveal manageable paths to an antiderivative, sometimes transforming an intimidating problem into a step-by-step journey worth understanding. Every piece simplified, represents a dimension of clarity being unfolded, making integration not only doable but also comprehensible.

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Most popular questions from this chapter

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$\begin{aligned} &\text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}=\\\ &\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{aligned}$$

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t)\) the Laplace transform is a new function \(F(s)\) defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ where we assume that \(s\) is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1}$$ Verify the following Laplace transforms, where a is a real number. $$f(t)=\cos a t \longrightarrow F(s)=\frac{s}{s^{2}+a^{2}}$$

Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). \(\int_{0}^{\pi / 2} \cos ^{9} x d x=\frac{128}{315}\)

Graph the integrands and then evaluate and compare the values of \(\int_{0}^{\infty} x e^{-x^{2}} d x\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\).

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0)\) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). Evaluate for \(\int \frac{\sqrt{x^{2}-1}}{x^{3}} d x,\) for \(x>1\) and for \(x<-1\)
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