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Chapter 7: Problem 23

Evaluate the integrals by making appropriate substitutions. $$\int \frac{\sin (5 / x)}{x^{2}} d x$$

Short Answer

Expert verified
The integral evaluates to \(\frac{1}{5}\cos\left(\frac{5}{x}\right) + C\).

Step by step solution

01

Identify the substitution

In this integral, the expression \(5/x\) in the sine function's argument suggests using substitution to simplify the expression. We can let \(u = 5/x\).
02

Differentiate the substitution

To replace \(dx\) in terms of \(du\), differentiate the substitution expression \(u = 5/x\). This gives us \(du = -\frac{5}{x^2} dx\).
03

Solve for dx

From the derivative \(du = -\frac{5}{x^2} dx\), solve for \(dx\). This gives \(dx = -\frac{x^2}{5} du\).
04

Substitute in the integral

Substitute \(u = 5/x\) and \(dx = -\frac{x^2}{5} du\) into the original integral: \[\int \frac{\sin \left( \frac{5}{x} \right) }{x^2} dx = \int \sin(u) \left(-\frac{1}{5}\right) du\].
05

Simplify the integral

We are left with the simpler integral \( -\frac{1}{5} \int \sin(u) \, du\).
06

Integrate

The integral of \(\sin(u)\) is \(-\cos(u)\), so we have:\[-\frac{1}{5}(-\cos(u)) + C = \frac{1}{5}\cos(u) + C\].
07

Back substitute

Replace \(u\) with \(5/x\) to get back to the original variable:\(\frac{1}{5}\cos\left(\frac{5}{x}\right) + C\).

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Key Concepts

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

Integration by Substitution
This technique is used in calculus to simplify difficult integrals. It involves changing the variable of integration to make the integral easier to solve. The key steps include choosing a substitution variable and differentiating it, substituting it into the original integral, and then integrating.
  • The goal is to find a function inside the integral that when replaced simplifies the process.
  • In our example, we noticed that the expression \(5/x\) inside the sine suggested using substitution.
  • By letting \(u = 5/x\), and differentiating, we transitioned the integral into a simpler form.

Remember, integration by substitution can transform complex integrals into more manageable tasks. It essentially reverses the chain rule used in differentiation and is invaluable for certain complex integrals.
Trigonometric Integrals
These types of integrals involve trigonometric functions like sine, cosine, tangent, and their inverses. Integrating such functions often requires special techniques or transformations.
  • To integrate trigonometric functions, familiarity with their derivatives is key. For example, the derivative of \( \cos(u) \) is \(-\sin(u)\).
  • Many times, trigonometric identities can help simplify integrals and make them more approachable.

In this exercise, the integral involves \( \sin(5/x) \). By transforming the integral with substitution, we reduced it to a standard form involving just \( \sin(u) \), which is straightforward to integrate. Using substitution allowed us to leverage known integral solutions of basic trigonometric functions without the complex argument inside the sine.
Differentiation
Differentiation is the process of finding the derivative of a function, which represents the rate of change of the function with respect to a variable. It's the inverse operation of integration, and often used in conjunction to solve integrals.
  • In the integration by substitution process, differentiation is a crucial step that provides a way to express \( dx \) in terms of the new variable.
  • For our exercise, from \( u = 5/x \), differentiating helped derive \( du = -\frac{5}{x^2} dx \), which was necessary for substitution.

While the main focus here was to simplify the integration process, understanding differentiation helps in recognizing how parts of a function interact, providing a necessary tool for handling substitutions effectively.

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Most popular questions from this chapter

(a) Use a graphing utility to generate the graph of $$f(x)=\frac{1}{100}(x+2)(x+1)(x-3)(x-5) $$ and use the graph to make a conjecture about the sign of the integral $$\int_{-2}^{5} f(x) d x$$ (b) Check your conjecture by evaluating the integral.

It was shown in the proof of the Mean-Value Theorem for Integrals that if \(f\) is continuous on \([a, b],\) and if \(m \leq f(x) \leq M\) on \([a, b],\) then $$m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)$$Isee \((8)] .\) These inequalities make it possible to obtain bounds on the size of a definite integral from bounds on the size of its integrand. This is illustrated. Find values of \(m\) and \(M\) such that \(m \leq x \sin x \leq M\) for \(0 \leq x \leq \pi,\) and use these values to find bounds on the value of the integral $$\int_{0}^{\pi} x \sin x d x$$

Find the area under the curve \(y=3\) sin \(x\) over the interval \([0,2 \pi / 3] .\) Sketch the region.

(a) The temperature of a 10 -m-long metal bar is \(15^{\circ} \mathrm{C}\) at one end and \(30^{\circ} \mathrm{C}\) at the other end. Assuming that the temperature increases linearly from the cooler end to the hotter end, what is the average temperature of the bar? (b) Explain why there must be a point on the bar where the temperature is the same as the average, and find it.

(a) Prove: If \(f\) is continuous on \([a, b],\) then $$\int_{a}^{b}\left[f(x)-f_{\text {ave }}\right] d x=0$$ (b) Does there exist a constant \(c \neq f_{\text {ave }}\) such that $$\int_{a}^{b}[f(x)-c] d x=0 ?$$
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