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Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
find the value.
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
have you tried using completing the square method?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
Looks like a candidate for trig substitution.
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
i do know the basic radical, but integrating radicals like that?.....oh thats confusing....
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
\[a=3,\ b=\sqrt{2}\]See what happens...
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
then what will be next?
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.0
Looks messy, but it really simplifies well.
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
looks messy. then what will be next.....
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
\[ \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)} \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
\[ = \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta) \]
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.0
I was simplifying part of it. It's still there.
 one year ago

Jnlucero Group TitleBest ResponseYou've already chosen the best response.0
so how will i change it into \[\sec \theta d \theta\]
 one year ago
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