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Chapter 8: Problem 23

Use integration tables to find the indefinite integral. \(\int e^{x} \arccos e^{x} d x\)

Short Answer

Expert verified
\(\int e^{x} \arccos e^{x} dx = e^{x} \arccos e^{x} + 2 \sinh^{-1} e^{x} + C \)

Step by step solution

01

Specify the Parts for Integration

We will apply the integration by parts formula: \(\int u dv = uv - \int v du \), where \(u\) and \(v\) are functions of \(x\). We'll choose \(u = \arccos e^{x}\) and \(dv = e^{x}dx\). After making these choices, we need to compute \(du\) and \(v\).
02

Compute \(du\) and \(v\)

Derive \(u\) to find \(du\). The derivative of \(\arccos x\) is \(-1/\sqrt{1-x^{2}}\), so \(du = -e^{-x}dx/\sqrt{1-e^{2x}}\). Integrate \(dv\) to find \(v\). The integral of \(e^{x}\) is \(e^{x}\) so that \(v = e^{x}\).
03

Apply the Integration by Parts Formula

Now we substitute our \(u\), \(v\), and \(du\) into the integration by parts formula: \(\int u dv = uv - \int v du\). After substitution, we have \(\int e^{x} \arccos e^{x} dx = e^{x} \arccos e^{x} - \int e^{x} (-e^{-x} dx / \sqrt{1 - e^{2x}})\). Further simplification leads to \( e^{x} \arccos e^{x} + \int dx / \sqrt{1 - e^{2x}} \).
04

Resolve Remaining Integral

The remaining integral can be solved by using trigonometric substitution. If we set \(e^{x} = \sinh t\), then \(dx = \cosh t dt\), and \(1 - e^{2x} = 1 - \sinh^{2} t = \cosh^{2} t\). Substituting back yields \(\int dx / \sqrt{1 - e^{2x}} = \int \cosh t dt = \sinh t = e^{x} - e^{-x}\). So our full solution is: \(e^{x} \arccos e^{x} + \sinh^{-1} e^{x} + C.\)
05

Simplify

We should simplify \(e^{x} - e^{-x}\) to \(2 \sinh (x)\). Now, putting everything together, we have: \(e^{x} \arccos e^{x} + 2 \sinh^{-1} e^{x} + C\).

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Key Concepts

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

Integration by Parts
When faced with an integral that is a product of two functions, the method of 'integration by parts' is a powerful tool. It's based on the product rule from differential calculus and can be applied to find indefinite integrals that are otherwise difficult to solve. The formula ewline \( \int u dv = uv - \int v du \) ewline is the backbone of this method, where one must wisely choose which part of the function to differentiate (\(u\)) and which to integrate (\(dv\)).
For the given exercise, the choice of \(u = \arccos e^{x}\) and \(dv = e^{x}dx\) is strategic because the resulting \(du\) and \(v\) lead to a simpler integral. After computing these values, we plug them back into the formula to progress stepwise towards the solution.
Trigonometric Substitution
Another valuable method for evaluating integrals is 'trigonometric substitution'. It's particularly helpful when dealing with integrals involving square roots of quadratic expressions. By substituting trigonometric functions for certain expressions, we can simplify the integral to a form that is easier to handle.
In the context of our exercise, after applying integration by parts, a trigonometric substitution is needed for the term \(\sqrt{1 - e^{2x}}\). By setting \(e^{x} = \sinh t\), we tap into the hyperbolic identity \(1 - \sinh^{2} t = \cosh^{2} t\), which simplifies the square root in our integral, allowing for a straightforward integration.
Inverse Trigonometric Functions
The inverse trigonometric functions, like \(\arccos x\), \(\arcsin x\), and \(\arctan x\), often appear when solving calculus problems. These functions provide angles whose trigonometric functions yield specific values.
In the initial integral \(\int e^{x} \arccos e^{x} dx\), we are looking for the function whose cosine gives us \(e^{x}\). The significance of inverse trigonometric functions becomes evident when evaluating our final simplified expression. The term \(\sinh^{-1} e^{x}\), which appears as a result of trigonometric substitution, is the inverse hyperbolic sine function, leading us to the final answer for the integral.
Understanding the behavior and differentiation rules for inverse trigonometric functions is crucial because they enable us to navigate through integrals that include these functions as part of the integrand. Recognizing them can prompt the use of specific techniques like integration by parts or substitutions that lead to an elegant solution.

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Most popular questions from this chapter

True or False? In Exercises \(85-88\) , determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f^{\prime}\) is continuous on \([0, \infty)\) and \(\lim _{x \rightarrow \infty} f(x)=0,\) then \(\int_{0}^{\infty} f^{\prime}(x) d x=-f(0)\)

Laplace Transforms Let \(f(t)\) be a function defined for all positive values of \(t .\) The Laplace Transform of \(f(t)\) is defined by $$F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t$$ when the improper integral exists. Laplace Transforms are used to solve differential equations. In Exercises \(95-102,\) find the Laplace Transform of the function. $$ f(t)=\sinh a t $$

Exploration Consider the integral $$\int_{0}^{\pi / 2} \frac{4}{1+(\tan x)^{n}} d x$$ where \(n\) is a positive integer. (a) Is the integral improper? Explain. (b) Use a graphing utility to graph the integrand for \(n=2,4\) \(8,\) and \(12 .\) (c) Use the graphs to approximate the integral as \(n \rightarrow \infty\) . (d) Use a computer algebra system to evaluate the integral for the values of \(n\) in part (b). Make a conjecture about the value of the integral for any positive integer \(n .\) Compare your results with your answer in part (c).

Proof Prove the following generalization of the Mean Value Theorem. If \(f\) is twice differentiable on the closed interval \([a, b],\) then $$f(b)-f(a)=f^{\prime}(a)(b-a)-\int_{a}^{b} f^{\prime \prime}(t)(t-b) d t$$

Evaluating a Limit Let \(f(x)=x+x \sin x\) and \(g(x)=x^{2}-4 .\) (a) Show that \(\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0\) (b) Show that \(\lim _{x \rightarrow \infty} f(x)=\infty\) and \(\lim _{x \rightarrow \infty} g(x)=\infty\) (c) Evaluate the limit $$\quad \lim _{x \rightarrow \infty} \frac{f^{\prime}(x)}{g^{\prime}(x)}$$ What do you notice? (d) Do your answers to parts (a) through (c) contradict L'Hopital's Rule? Explain your reasoning.
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