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Chapter 6: Problem 32

Determine each indefinite integral. $$\int \operatorname{sech}^{2} x \tanh x d x$$

Short Answer

Expert verified
Based on the given step by step solution, the integral of the function is: $$\int \operatorname{sech}^{2}(x)\tanh(x) \, d x = \frac{1}{2}\tanh^2(x) + C$$

Step by step solution

01

Define the substitution

Let's define a substitution \(u = \tanh(x)\), so we can replace \(\operatorname{sech}^{2}(x)\tanh(x)\) in the integral.
02

Find the differential

Now we need to find \(\frac{d u}{d x}\) to make the substitution. Differentiate \(u = \tanh(x)\) with respect to \(x\): $$\frac{d u}{d x} = \frac{d (\tanh x)}{d x}$$ We know that the derivative of \(\tanh(x)\) is \(\operatorname{sech}^2(x)\). So, $$\frac{du}{dx} = \operatorname{sech}^2(x)$$ Now, we can find \(dx\) in terms of \(du\): $$dx = \frac{1}{\operatorname{sech}^2(x)} \, du$$
03

Rewrite the integral using the substitution

Substitute \(u=\tanh(x)\) and \(dx\) in the original integral: $$\int \operatorname{sech}^{2}(x) \tanh(x) \, d x = \int \operatorname{sech}^{2}(x) \cdot u \cdot \frac{1}{\operatorname{sech}^2(x)} \, du$$ Now, we can cancel the \(\operatorname{sech}^2(x)\) terms, resulting in: $$\int u \, du$$
04

Integrate with respect to \(u\)

Integrate \(\int u \, du\): $$\frac{1}{2}u^2 + C$$
05

Replace \(u\) with \(\tanh(x)\)

Replace \(u\) with \(\tanh(x)\) to express the result in terms of the original variable \(x\): $$\frac{1}{2}\tanh^2(x)+C$$
06

Write the final result

The integral of \(\operatorname{sech}^{2}(x)\tanh(x)\) with respect to \(x\) is: $$\int \operatorname{sech}^{2} x \tanh x d x = \frac{1}{2}\tanh^2(x) + C$$

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Key Concepts

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

Indefinite Integrals
Indefinite integrals are fundamental in calculus, describing the family of functions whose derivative is the original function we are integrating. They are expressed generally in the form \[ \int f(x) \, dx = F(x) + C \]where \( F(x) \) is the antiderivative of \( f(x) \), and \( C \) is the constant of integration.
Unlike definite integrals, indefinite integrals do not have specific start and endpoint values.
The constant \( C \) is crucial as it accounts for the family of possible functions differing only by a constant, which all have the same derivative.
  • The role of \( C \) ensures that every potential vertical shift of the antiderivative is captured.
  • Indefinite integrals are closely related to finding areas under curves, but in a more general sense.
Understanding the indefinite integral helps in determining general solutions to differential equations and forms the basis for solving problems involving accumulation and changes.
Substitution Method
The substitution method is a common technique in calculus used to simplify the process of finding antiderivatives, especially when dealing with composite functions. The main idea is to make the integral easier to evaluate by changing variables.
In our case, we chose \( u = \tanh(x) \). This effectively transformed the original complex integral into a simpler one by:
  • Identifying a part of the integrand that could be expressed as a function of a new variable \( u \).
  • Finding the differential \( du \) to substitute in for \( dx \).
  • Translating the entire integrand terms in terms of \( u \) and \( du \).
This method facilitates easier computation, and after integration is performed in terms of \( u \), we revert back to the original variable, ensuring the final solution remains in the context of the original problem.
Hyperbolic Functions
Hyperbolic functions often appear in calculus problems related to integration and differentiation due to their unique properties, which are similar yet distinct compared to trigonometric functions.
The key hyperbolic functions used here are \( \tanh(x) \) and \( \operatorname{sech}(x) \):
  • The hyperbolic tangent \( \tanh(x) \) is defined as \( \frac{\sinh(x)}{\cosh(x)} \), mimicking the coastline ratio.
  • The hyperbolic secant \( \operatorname{sech}(x) \) is the reciprocal of \( \cosh(x) \), much like trigonometric secant, and is given by \( \frac{1}{\cosh(x)} \).

They are critical in this exercise because:
  • The derivative of \( \tanh(x) \) is \( \operatorname{sech}^2(x) \), which simplifies part of our integration process.
  • These functions often arise in solutions to differential equations and in modeling real-world phenomena involving growth and decay.
Understanding their properties and derivatives makes handling integrals involving hyperbolic functions much more manageable in complex calculations.

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Most popular questions from this chapter

A 30-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and the chain has a density of \(5 \mathrm{kg} / \mathrm{m}\). a. How much work is required to wind the entire chain onto the cylinder using the winch? b. How much work is required to wind the chain onto the cylinder if a \(50-\mathrm{kg}\) block is attached to the end of the chain?

Find the volume of the solid generated in the following situations. The region \(R\) in the first quadrant bounded by the graphs of \(y=x\) and \(y=1+\frac{x}{2}\) is revolved about the line \(y=3\).

Use a calculator to evaluate each expression or state that the value does not exist. Report answers accurate to four decimal places. a. \(\coth 4\) b. \(\tanh ^{-1} 2\) c. \(\operatorname{csch}^{-1} 5\) d. \(\left.\operatorname{csch} x\right|_{1 / 2} ^{2}\) e. \(\left.\ln \left|\tanh \left(\frac{x}{2}\right)\right|\right|_{1} ^{10}\) f. \(\left.\tan ^{-1}(\sinh x)\right|_{-3} ^{3}\) g. \(\left.\frac{1}{4} \operatorname{coth}^{-1} \frac{x}{4}\right|_{20} ^{36}\)

Suppose the acceleration of an object moving along a line is given by \(a(t)=-k v(t),\) where \(k\) is a positive constant and \(v\) is the object's velocity. Assume that the initial velocity and position are given by \(v(0)=10\) and \(s(0)=0,\) respectively. a. Use \(a(t)=v^{\prime}(t)\) to find the velocity of the object as a function of time. b. Use \(v(t)=s^{\prime}(t)\) to find the position of the object as a function of time. c. Use the fact that \(d v / d t=(d v / d s)(d s / d t)\) (by the Chain Rule) to find the velocity as a function of position.

Show that \(\cosh ^{-1}(\cosh x)=|x|\) by using the formula \(\cosh ^{-1} t=\ln (t+\sqrt{t^{2}-1})\) and by considering the cases \(x \geq 0\) and \(x<0.\)
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