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Chapter 7: Problem 24

Use a table of integrals to determine the following indefinite integrals. These integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table. $$\int \sqrt{x^{2}-8 x} d x, x>8$$

Short Answer

Expert verified
The indefinite integral of the given function is: $$\int \sqrt{x^{2}-8 x} dx = 2x - 8 + C$$

Step by step solution

01

Complete the square

Let's first complete the square for the expression inside the square root: $$x^2 - 8x = (x - 4)^2 - 16$$ Now, we rewrite the given integral using this new expression: $$\int \sqrt{(x - 4)^2 - 16} dx$$
02

Perform a change of variables

To simplify the integral, let's try a trigonometric substitution. We'll let $$x - 4 = 4\sec(\theta)$$. This substitution is helpful as it will allow us to use the identity $$\sec^2(\theta) - 1 = \tan^2(\theta)$$ which simplifies the expression inside the square root. Differentiating with respect to x we get $$dx = 4\sec(\theta)\tan(\theta) d\theta$$. Now, let's substitute the variables back into the integral: $$\int \sqrt{(4\sec(\theta))^2 - 16}(4\sec(\theta)\tan(\theta) d\theta)$$
03

Simplify the expression and solve the integral

Now, we can simplify the expression inside the square root and rearrange the factors of the integral: $$\int 4\sec(\theta)\tan(\theta) \sqrt{16\sec^2(\theta) - 16} d\theta$$ Notice that 16 can be factored out of the square root: $$\int 4\sec(\theta)\tan(\theta) (4\sqrt{\sec^2(\theta)-1}) d\theta$$ Next, we apply the identity $$\sec^2(\theta) - 1 = \tan^2(\theta)$$: $$\int 4\sec(\theta)\tan(\theta) (4\sqrt{\tan^2(\theta)}) d\theta$$ Now, we can simplify square root and solve the integral: $$\int 16\sec(\theta)\tan^2(\theta) d\theta$$ $$= \int (\frac{16}{2}\sec(\theta))(2\sec(\theta)\tan(\theta)) d\theta$$ $$= 8 \int \sec(\theta) d(\sec(\theta))$$ So the new integral becomes, $$8\sec(\theta) + C$$
04

Reverse the substitution

Now, we need to reverse the substitution and find the original function in terms of x. From previous substitution, we know that $$x - 4 = 4\sec(\theta)$$. Therefore, $$\sec(\theta) = \frac{x - 4}{4}$$. Substituting this into our current integral result, we get: $$8\left(\frac{x - 4}{4}\right) + C$$
05

Simplify the expression

Finally, let's simplify the expression: $$2(x - 4) + C$$ $$2x - 8 + C$$ So, the indefinite integral of the given function is: $$\int \sqrt{x^{2}-8 x} dx = 2x - 8 + C$$

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Most popular questions from this chapter

The cycloid is the curve traced by a point on the rim of a rolling wheel. Imagine a wire shaped like an inverted cycloid (see figure). A bead sliding down this wire without friction has some remarkable properties. Among all wire shapes, the cycloid is the shape that produces the fastest descent time. It can be shown that the descent time between any two points \(0 \leq a \leq b \leq \pi\) on the curve is $$\text { descent time }=\int_{a}^{b} \sqrt{\frac{1-\cos t}{g(\cos a-\cos t)}} d t$$ where \(g\) is the acceleration due to gravity, \(t=0\) corresponds to the top of the wire, and \(t=\pi\) corresponds to the lowest point on the wire. a. Find the descent time on the interval \([a, b]\) by making the substitution \(u=\cos t\) b. Show that when \(b=\pi\), the descent time is the same for all values of \(a ;\) that is, the descent time to the bottom of the wire is the same for all starting points.

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt[4]{x+2}+1} ; x+2=u^{4}$$

Determine whether the following statements are true and give an explanation or counterexample. a. To evaluate \(\int \frac{4 x^{6}}{x^{4}+3 x^{2}} d x\), the first step is to find the partial fraction decomposition of the integrand. b. The easiest way to evaluate \(\int \frac{6 x+1}{3 x^{2}+x} d x\) is with a partial fraction decomposition of the integrand. c. The rational function \(f(x)=\frac{1}{x^{2}-13 x+42}\) has an irreducible quadratic denominator. d. The rational function \(f(x)=\frac{1}{x^{2}-13 x+43}\) has an irreducible quadratic denominator.

Use the indicated substitution to convert the given integral to an integral of a rational function. Evaluate the resulting integral. $$\int \frac{d x}{\sqrt{1+\sqrt{x}}} ; x=\left(u^{2}-1\right)^{2}$$

On the interval \([0,2],\) the graphs of \(f(x)=x^{2} / 3\) and \(g(x)=x^{2}\left(9-x^{2}\right)^{-1 / 2}\) have similar shapes. a. Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval [0,2] b. Find the area of the region bounded by the graph of \(g\) and the \(x\) -axis on the interval [0,2] c. Which region has the greater area?
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