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Chapter 7: Problem 20

Evaluate the following integrals. $$\int x \sec ^{-1} x d x, x \geq 1$$

Short Answer

Expert verified
Question: Evaluate the integral of the function $$f(x) = x \sec^{-1}(x)$$, with the domain $$x \geq 1$$. Answer: $$\frac{1}{2}x^2\sec^{-1}(x) - \frac{1}{2}\sqrt{x^2-1} + C$$

Step by step solution

01

Apply integration by parts

To solve the integral, we shall use integration by parts. Recall the integration by parts formula: $$\int u dv = uv - \int v du$$ For this problem, let's choose: $$u = \sec^{-1}(x)$$ $$dv = x dx$$
02

Differentiate \(u\) and integrate \(dv\)

In our case, we need to find the derivatives of the chosen function \(u\), and the integral of the chosen term \(dv\). We have: $$du = \frac{d(\sec^{-1}(x))}{dx} dx = \frac{dx}{x\sqrt{x^2-1}}$$ $$v = \int x dx = \frac{1}{2}x^2$$
03

Substitute \(u\), \(v\), \(du\), and \(dv\) in the integration by parts formula

Now substitute the values of \(u\), \(v\), \(du\), and \(dv\) back into the integration by parts formula: $$\int x \sec^{-1}(x) dx = \left(\frac{1}{2}x^2\right)\sec^{-1}(x) - \int \left(\frac{1}{2}x^2\right) \left(\frac{dx}{x\sqrt{x^2-1}}\right)$$
04

Simplify the integral

Simplify the remaining integral: $$\int x \sec^{-1}(x) dx = \frac{1}{2}x^2\sec^{-1}(x) - \frac{1}{2}\int \frac{x^2}{x\sqrt{x^2-1}} dx = \frac{1}{2}x^2\sec^{-1}(x) - \frac{1}{2}\int \frac{x}{\sqrt{x^2-1}} dx$$
05

Perform substitution

To evaluate the remaining integral, perform the substitution: $$t = x^2 - 1$$ $$dt = 2x dx$$ The integral becomes: $$-\frac{1}{2}\int \frac{x dx}{\sqrt{x^2-1}} = -\frac{1}{4}\int \frac{dt}{\sqrt{t}}$$
06

Evaluate the remaining integral

Now, evaluate the remaining integral: $$-\frac{1}{4}\int\frac{dt}{\sqrt{t}} = -\frac{1}{4}\int t^{-1/2} dt = -\frac{1}{4}(2t^{1/2}) = -\frac{1}{2}\sqrt{t} + C$$
07

Substitute back and simplify

Finally, substitute back the variable \(t\): $$-\frac{1}{2}\sqrt{t} + C = -\frac{1}{2}\sqrt{x^2-1} + C$$
08

Combine all terms

Combine all terms to get the final antiderivative of the function: $$\int x \sec^{-1}(x) dx = \frac{1}{2}x^2\sec^{-1}(x) - \frac{1}{2}\sqrt{x^2-1} + C$$ Thus, the integral of the given function is: $$\int x \sec^{-1}(x) dx = \frac{1}{2}x^2\sec^{-1}(x) - \frac{1}{2}\sqrt{x^2-1} + C$$

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Most popular questions from this chapter

\(A\) 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. $$f(t)=t \longrightarrow F(s)=\frac{1}{s^{2}}$$

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \sqrt{e^{x}+1} d x \text { (Hint: Let } u=\sqrt{e^{x}+1}$$

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

Challenge Show that with the change of variables \(u=\sqrt{\tan x}\) the integral \(\int \sqrt{\tan x} d x\) can be converted to an integral amenabl to partial fractions. Evaluate \(\int_{0}^{\pi / 4} \sqrt{\tan x} d x\)

Let \(I_{n}=\int x^{n} e^{-x^{2}} d x,\) where \(n\) is a nonnegative integer. a. \(I_{0}=\int e^{-x^{2}} d x\) cannot be expressed in terms of elementary functions. Evaluate \(I_{1}\). b. Use integration by parts to evaluate \(I_{3}\). c. Use integration by parts and the result of part (b) to evaluate \(I_{5}\). d. Show that, in general, if \(n\) is odd, then \(I_{n}=-\frac{1}{2} e^{-x^{2}} p_{n-1}(x)\) where \(p_{n-1}\) is a polynomial of degree \(n-1\). e. Argue that if \(n\) is even, then \(I_{n}\) cannot be expressed in terms of elementary functions.
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