/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Evaluate the integral. $$\int ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integral. $$\int e^{-x} \tan \left(e^{-x}\right) d x$$

Short Answer

Expert verified
\( \ln |\cos(e^{-x})| + C \)

Step by step solution

01

Substitution

We start by using substitution. Let \( u = e^{-x} \). Then, the derivative \( du = -e^{-x} dx = -u dx \), so \( dx = -\frac{du}{u} \). This substitution transforms the integral.
02

Transform the Integral

Substitute \( u \) and \( dx \) in the integral: \[ \int e^{-x} \tan(e^{-x}) \, dx = \int u \tan(u) \left(-\frac{du}{u}\right) = -\int \tan(u) \, du \] after canceling \( u \) out.
03

Integrate

We now need to integrate \( -\int \tan(u) \, du \). The integral of \( \tan(u) \) is \( -\ln |\cos(u)| + C \). Thus, we get: \[ -\left( -\ln |\cos(u)| \right) = \ln |\cos(u)| + C \]
04

Substitute Back

Substitute \( u = e^{-x} \) back into the expression: \( \ln |\cos(e^{-x})| + C \). Thus, the evaluated integral is: \[ \ln |\cos(e^{-x})| + C \]

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Key Concepts

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

Substitution Method
The Substitution Method is a powerful tool in calculus that simplifies the process of integration by changing variables. Essentially, it's about finding a substitution that makes an integral easier to evaluate. The basic idea is to identify a part of the integral that can be set as a new variable, often denoted as \( u \). Once you've identified this substitution, the next step is to express the differential \( dx \) in terms of \( du \). This is achieved by differentiating \( u \) with respect to \( x \) to find \( du \).
  • Start by letting \( u = g(x) \), where \( g(x) \) is a part of the integrand.
  • Differentiate to find \( du = g'(x) dx \).
  • Express \( dx \) in terms of \( du \) and \( g(x) \).
By substituting \( u \) and \( dx \) back into the integral, you transform the original integral into a simpler one that's easier to solve. After integration, substitute the original expression back for \( u \). This method can look complex, but with practice, it becomes a straightforward step-by-step process.
Exponential Functions
Exponential functions are a class of mathematical functions denoted by \( e^x \), where \( e \) (approximately 2.718) is the base of the natural logarithm. They are unique because their rate of growth is proportional to their size, making them widely applicable in various fields such as mathematics, physics, and economics.In the given problem, the exponential function is \( e^{-x} \). This form is notable for its characteristic of decaying rather than growing, which can model numerous phenomena such as radioactive decay or cooling processes.
  • Exponential decay: as \( x \) increases, \( e^{-x} \) approaches zero.
  • Simplifies to manageable limits and differentiations with the substitution method.
Handling exponential functions often involves transformations like those seen in substitution methods. Substituting \( e^{-x} \) with a new variable simplifies the integration process, as seen in the given problem, turning a potentially complex integral into a much simpler one to evaluate.
Trigonometric Integration
Trigonometric Integration involves integrating functions that include trigonometric expressions such as \( \sin(x) \), \( \cos(x) \), or \( \tan(x) \). These functions are periodic and oscillate, which makes their integration sometimes intricate yet fascinating.In the example problem, the focus was on integrating \( \tan(u) \). The function \( \tan(x) \), or tangent, can be set as \( \frac{\sin(x)}{\cos(x)} \), making its integration a bit more complex than sine or cosine alone.
  • A standard approach for \( \tan(x) \) is recognizing its integral as \( -\ln|\cos(x)|+C \).
  • Use identities and algebraic manipulation for simplifying integrals.
Because of periodic properties, trigonometric integrals often utilize identities or transformations. Identifying patterns and using known derivatives and integrals of these functions is key. By integrating \( \tan(u) \), we exploited the relationship between tangent and cosine to simplify and solve the integral efficiently.

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Most popular questions from this chapter

Medication can be administered to a patient using a variety of methods. For a given method, let \(c(t)\) denote the concentration of medication in the patient's bloodstream (measured in \(\mathrm{mg} / \mathrm{L}\) ) \(t\) hours after the dose is given. The area under the curve \(c=c(t)\) over the time interval \([0,+\infty)\) indicates the "availability" of the medication for the patient's body. Determine which method provides the greater availability. Method \(1: c_{1}(t)=6\left(e^{-0.4 t}-e^{-1.3 t}\right)\) Method \(2: c_{2}(t)=5\left(e^{-0.4 t}-e^{-3 t}\right)\)

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Suppose that during a period \(t_{0} \leq t \leq t_{1}\) years, a company has a continuous income stream at a rate of \(I(t)\) dollars per year at time \(t\) and that this income is invested at an annual rate of \(r \%,\) compounded continuously. The value (in dollars) of this income stream at the end of the time period \(t_{0} \leq t \leq t_{1},\) called the stream's future value, can be calculated using $$F V=\int_{t_{0}}^{t_{1}} I(t) e^{r\left(t_{1}-t\right)} d t$$ The present value (in dollars) of the income stream is given by $$P V=\int_{t_{0}}^{t_{1}} I(t) e^{-r\left(t-t_{0}\right)} d t$$ The present value is the amount that, if put in the bank at time \(t=t_{0}\) at \(r \%\) compounded continuously, with no additional deposits, would result in a balance of \(F V\) dollars at time \(t=t_{1}\) That is, $$F V=P V \times e^{r\left(t_{1}-t_{0}\right)}$$ In each exercise, (a) find the future value \(F V\) for the given income stream \(I(t)\) and interest rate \(r\) and time period \(t_{0} \leq t \leq t_{1}\) (b) find the present value \(P V\) of the income stream over the time period; and (c) verify that \(F V\) and \(P V\) satisfy the relationship given above. $$I(t)=2000 t+400 e^{-t} ; r=8 \% ; 0 \leq t \leq 25$$

Find the volume of the solid that results when the region enclosed by \(y=\cos x, y=\sin x, x=0,\) and \(x=\pi / 4\) is revolved about the \(x\) -axis.

Determine whether the statement is true or false. Explain your answer. \(\int_{1}^{2} \frac{1}{x(x-3)} d x\) is an improper integral.

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