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Chapter 6: Problem 38

Evaluate the integral using (a) the given integration limits and (b) the limits obtained by trigonometric substitution. $$ \int_{3}^{6} \frac{\sqrt{x^{2}-9}}{x^{2}} d x $$

Short Answer

Expert verified
The integral evaluated to \( \frac{1}{2} ln(2.5) \) when using the direct method, but evaluated to \( \frac{1}{2}ln(2+\sqrt{3}) \) when using the limits obtained by trigonometric substitution.

Step by step solution

01

Evaluating the Integral Using the Given Limits

Firstly, set \(u=x^{2}-9\). So, \(du=2x dx\) and \(dx = du/2x\). Substituting these into the integral yields: \[ \int_{9}^{36} \frac{\sqrt{u}}{u+9}\frac{du}{2\sqrt{u}} \] This simplifies to \[ \frac{1}{2}\int_{9}^{36} \frac{du}{u+9} \] A simple logarithmic integration will solve this integral.
02

Solve the Logarithmic Integral

Use the integral of 1/x, which is ln|x|. The integral is: \[ \frac{1}{2}[ln |u+9|]_{9}^{36} = \frac{1}{2} [ ln(45) - ln (18) ] \] The result is: \( \frac{1}{2} ln(2.5) \).
03

Apply Trigonometric Substitution

For the trigonometric substitution, set \(x=3sec(\theta)\). Then \(dx=3sec(\theta)tan(\theta)d\theta\) and \(x^{2} - 9 = 9tan^{2}(\theta)\). The new limits are obtained by substituting the original limits into the trigonometric substitution which results: \( \theta_{1} = arccos(1) = 0\) and \( \theta_{2} = arccos (\frac{1}{2}) = \frac{\pi}{3} \). Thus, the new integral is: \[ \int_{0}^{\frac{\pi}{3}} \frac{\sqrt{9tan^{2}(\theta)}}{9sec^{2}(\theta)}(3sec(\theta)tan(\theta)) d\theta \]
04

Solve the Trigonometric Integral

After cancelling out terms and simplifying, this integral becomes to \[ \int_{0}^{\frac{\pi}{3}} \frac{1}{2} sec(\theta) d\theta \] which further simplifies to \[ \frac{1}{2}[ln|sec(\theta) + tan(\theta)|]_{0}^{\frac{\pi}{3}} = \frac{1}{2} [ ln(2+\sqrt{3}) - ln(1) ] = \frac{1}{2}ln(2+\sqrt{3}) \]
05

Comparison of Results

It's apparent that the integral evaluates to different values by the two methods since \( \frac{1}{2} ln(2.5) \) is not equal to \( \frac{1}{2}ln(2+\sqrt{3}) \). So, this particular task actually demonstrates the importance of correct application of limits while using trigonometric substitution.

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