anonymous
  • anonymous
Please help:
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]
anonymous
  • anonymous
find the value.
lgbasallote
  • lgbasallote
have you tried using completing the square method?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Looks like a candidate for trig substitution.
anonymous
  • anonymous
i do know the basic radical, but integrating radicals like that?.....oh thats confusing....
lgbasallote
  • lgbasallote
you 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
  • anonymous
Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).
anonymous
  • anonymous
\[a=3,\ b=\sqrt{2}\]See what happens...
anonymous
  • anonymous
then what will be next?
anonymous
  • anonymous
Well 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
  • anonymous
Looks messy, but it really simplifies well.
anonymous
  • anonymous
looks messy. then what will be next.....
anonymous
  • anonymous
\[ \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)} \]
anonymous
  • anonymous
\[ = \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta) \]
anonymous
  • anonymous
what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?
anonymous
  • anonymous
I was simplifying part of it. It's still there.
anonymous
  • anonymous
ok......
anonymous
  • anonymous
so how will i change it into \[\sec \theta d \theta\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.