- #1

ago01

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- Homework Statement
- Integrate ##\frac{1}{\sqrt{x^2-49}}dx##

- Relevant Equations
- Trig sub, secant.

I was doing a practice quiz and got the following integral:

##\int \frac{1}{\sqrt{x^2-49}}dx##

Following normal trig sub. Using ##b^2 = c^2 - a^2## we find ##\sec{\theta} = \frac{x}{5}## and so ##x = 5\sec{\theta}## and ##dx = 5\sec{\theta}\tan{\theta}d\theta##

Then

## \int \frac{1}{\sqrt{x^2-49}}dx = 5 \int \frac{\sec{\theta}\tan{\theta}d\theta}{\tan{\theta}}##

Then after simplification:

##\int \sec{\theta}d\theta = ln|\sec{\theta} + \tan{\theta}| + C##

and solving for theta using the triangle:

## ln|\frac{x}{7} + \frac{\sqrt{x^2-49}}{7}| + C ##

The autograder marked me wrong. The correct answer was:

##ln|\frac{1}{7}(\sqrt{x^2-49} + x)| + C##

Which looks to be the same to me. I cannot see the difference. Even the plots look the same. I must be forgetting a property of logs that makes these distinct. Can you help me?

##\int \frac{1}{\sqrt{x^2-49}}dx##

Following normal trig sub. Using ##b^2 = c^2 - a^2## we find ##\sec{\theta} = \frac{x}{5}## and so ##x = 5\sec{\theta}## and ##dx = 5\sec{\theta}\tan{\theta}d\theta##

Then

## \int \frac{1}{\sqrt{x^2-49}}dx = 5 \int \frac{\sec{\theta}\tan{\theta}d\theta}{\tan{\theta}}##

Then after simplification:

##\int \sec{\theta}d\theta = ln|\sec{\theta} + \tan{\theta}| + C##

and solving for theta using the triangle:

## ln|\frac{x}{7} + \frac{\sqrt{x^2-49}}{7}| + C ##

The autograder marked me wrong. The correct answer was:

##ln|\frac{1}{7}(\sqrt{x^2-49} + x)| + C##

Which looks to be the same to me. I cannot see the difference. Even the plots look the same. I must be forgetting a property of logs that makes these distinct. Can you help me?