A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Please help:
anonymous
 3 years ago
Please help:

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0have you tried using completing the square method?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Looks like a candidate for trig substitution.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i do know the basic radical, but integrating radicals like that?.....oh thats confusing....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you should use completing the square first....that would give you an integral that looks like \[\int \sqrt{x^2  a^2}dx\] does that look familiar?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[a=3,\ b=\sqrt{2}\]See what happens...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then what will be next?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well if you did the substitution, you'd get \[ x=\frac{3}{\sqrt{2}}\tan(\theta)\quad dx = \frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \] \[ \Large \int_1^2 \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2}\frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Looks messy, but it really simplifies well.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0looks messy. then what will be next.....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[ = \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta) \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was simplifying part of it. It's still there.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so how will i change it into \[\sec \theta d \theta\]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.