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Jnlucero
 3 years ago
Please help:
Jnlucero
 3 years ago
Please help:

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

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

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

Jnlucero
 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....

lgbasallote
 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?

wio
 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 \(x=\frac{a}{b}\tan(\theta)\).

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

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

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

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

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

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

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

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

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

Jnlucero
 3 years ago
Best ResponseYou've already chosen the best response.0so how will i change it into \[\sec \theta d \theta\]
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