/*! 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 15 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 15

Evaluate the following integrals. $$\int \frac{\sqrt{9-x^{2}}}{x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral of the function \(\frac{\sqrt{9-x^{2}}}{x}\). Answer: \(\int \frac{\sqrt{9-x^{2}}}{x} d x = -\frac{(9-x^2)\sqrt{9-x^2}}{27}+C\)

Step by step solution

01

Trigonometric Substitution

We can use the trigonometric substitution \(x = 3\sin(\theta)\) with \(d x = 3\cos(\theta)d\theta\). The integral then becomes: $$\int \frac{\sqrt{9-(3\sin(\theta))^2}}{3\sin(\theta)}(3\cos(\theta)d\theta)$$
02

Simplify the integrand

The integrand can now be simplified as follows: $$\int \frac{\sqrt{9-9\sin^2(\theta)}}{3\sin(\theta)}(3\cos(\theta)d\theta)$$ $$=\int \frac{\sqrt{9(1-\sin^2(\theta))}}{3\sin(\theta)}(3\cos(\theta)d\theta)$$ $$=\int \frac{3\sqrt{1-\sin^2(\theta)}}{\sin(\theta)}(\cos(\theta)d\theta)$$ We know that \(\cos^2(\theta) = 1 - \sin^2(\theta)\), so we can simplify the integrand further: $$=\int \frac{3\cos(\theta)}{\sin(\theta)}(\cos(\theta)d\theta)$$
03

Multiply and integrate

Next, we'll multiply the remaining functions, then integrate: $$=\int \frac{3\cos^2(\theta)}{\sin(\theta)}d\theta$$ Now, let's perform the integration with respect to \(\theta\): $$=3\int \frac{\cos^2(\theta)}{\sin(\theta)}d\theta$$ At this point, we perform another substitution. Let \(u = \cos(\theta)\) and \(d u = -\sin(\theta)d\theta\). We then get: $$=-3\int u^2d u$$
04

Integrate and reverse substitution

We can now integrate the expression: $$=-3\int u^2d u = -u^3 + C = -(\cos(\theta))^3 + C$$ Now, we need to reverse the substitutions to write the answer in terms of \(x\). We know that \(x = 3\sin(\theta)\), so \(\sin(\theta) = \frac{x}{3}\), and using the Pythagorean identity, we find \(\cos(\theta) = \frac{\sqrt{9-x^2}}{3}\). Substituting back, we get: $$=-\left(\frac{\sqrt{9-x^2}}{3}\right)^3 + C$$
05

Simplify the result

Finally, we can simplify our result: $$=-\frac{(9-x^2)\sqrt{9-x^2}}{27}+C$$ So the final solution is: $$\int \frac{\sqrt{9-x^{2}}}{x} d x = -\frac{(9-x^2)\sqrt{9-x^2}}{27}+C$$

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Most popular questions from this chapter

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

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