- #1
ago01
- 46
- 8
- 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?