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Use integral tables to evaluate the integral. \( \int \frac{\sqrt{\mathrm{x}^{2}-16}}{\mathrm{x}} \mathrm{dx} \) Click here to view page 1 of the Table of Integrals. Click he

Ask by May Harris. in the United States
Jan 24,2025

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Tutor-Verified Answer

Answer

The integral evaluates to \( \sqrt{x^2 - 16} - 4 \sec^{-1}\left(\frac{x}{4}\right) + C \).

Solution

To evaluate the integral \[ I = \int \frac{\sqrt{x^2 - 16}}{x} \, dx, \] we can use a substitution method. ### Step 1: Substitution Let's use the substitution \( x = 4 \sec(\theta) \). Then, we have: \[ dx = 4 \sec(\theta) \tan(\theta) \, d\theta. \] ### Step 2: Transform the integral Now, we can rewrite the integral in terms of \(\theta\): 1. Calculate \( \sqrt{x^2 - 16} \): \[ \sqrt{x^2 - 16} = \sqrt{(4 \sec(\theta))^2 - 16} = \sqrt{16 \sec^2(\theta) - 16} = \sqrt{16(\sec^2(\theta) - 1)} = \sqrt{16 \tan^2(\theta)} = 4 \tan(\theta). \] 2. Substitute into the integral: \[ I = \int \frac{4 \tan(\theta)}{4 \sec(\theta)} \cdot 4 \sec(\theta) \tan(\theta) \, d\theta = \int 4 \tan^2(\theta) \, d\theta. \] ### Step 3: Simplify the integral Using the identity \( \tan^2(\theta) = \sec^2(\theta) - 1 \), we can rewrite the integral: \[ I = 4 \int \tan^2(\theta) \, d\theta = 4 \int (\sec^2(\theta) - 1) \, d\theta = 4 \left( \int \sec^2(\theta) \, d\theta - \int 1 \, d\theta \right). \] ### Step 4: Evaluate the integrals The integrals can be evaluated as follows: 1. \( \int \sec^2(\theta) \, d\theta = \tan(\theta) + C_1 \). 2. \( \int 1 \, d\theta = \theta + C_2 \). Thus, \[ I = 4 \left( \tan(\theta) - \theta \right) + C. \] ### Step 5: Back-substitution Now we need to convert back to \(x\): 1. From our substitution \( x = 4 \sec(\theta) \), we have \( \sec(\theta) = \frac{x}{4} \) and thus \( \tan(\theta) = \sqrt{\sec^2(\theta) - 1} = \sqrt{\left(\frac{x}{4}\right)^2 - 1} = \frac{\sqrt{x^2 - 16}}{4} \). 2. Also, \( \theta = \sec^{-1}\left(\frac{x}{4}\right) \). Putting it all together, we have: \[ I = 4 \left( \frac{\sqrt{x^2 - 16}}{4} - \sec^{-1}\left(\frac{x}{4}\right) \right) + C = \sqrt{x^2 - 16} - 4 \sec^{-1}\left(\frac{x}{4}\right) + C. \] ### Final Answer Thus, the evaluated integral is: \[ \int \frac{\sqrt{x^2 - 16}}{x} \, dx = \sqrt{x^2 - 16} - 4 \sec^{-1}\left(\frac{x}{4}\right) + C. \]

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Bonus Knowledge

To evaluate the integral \( \int \frac{\sqrt{x^2 - 16}}{x} \, dx \), you can use a trigonometric substitution. Set \( x = 4\sec(\theta) \), which transforms the integral into a form that can be simplified. The differential \( dx \) then becomes \( 4\sec(\theta)\tan(\theta) \, d\theta \), allowing you to replace \( \sqrt{x^2 - 16} \) and \( x \) accordingly. This results in a nicer integral involving trigonometric functions that can be integrated easily. Once you perform the substitution and integrate, don't forget to back-substitute in terms of \( x \) to express your final answer in the original variable. This is a fun way to see how trigonometric identities play a role in simplifying what initially looks like a complicated integral!

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