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Jnlucero
Please help:
\[\int\limits_{1}^{3}\sqrt{2x^2+9} dx\]
have you tried using completing the square method?
Looks like a candidate for trig substitution.
i do know the basic radical, but integrating radicals like that?.....oh thats confusing....
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?
Thinks of the form \(\sqrt{a^2+b^2x^2} \) can be substituted with \(x=\frac{a}{b}\tan(\theta)\).
\[a=3,\ b=\sqrt{2}\]See what happens...
then what will be next?
Well if you did the substitution, you'd get \[ x=\frac{3}{\sqrt{2}}\tan(\theta)\quad dx = \frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \] \[ \Large \int_1^2 \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2}\frac{3}{\sqrt{2}}\sec^2(\theta)d\theta \]
Looks messy, but it really simplifies well.
looks messy. then what will be next.....
\[ \sqrt{\sqrt{2}^2\left(\frac{3}{\sqrt{2}}\tan(\theta)\right)^2+3^2} = \sqrt{3^2\tan^3(\theta)+3^2} = \sqrt{3^2(\tan^2(\theta)+1)} \]
\[ = \sqrt{3^2\sec^2(\theta)} = 3\sec(\theta) \]
what happened to \[\frac{ 3 }{ \sqrt{2} }\sec^2\theta d \theta\]?
I was simplifying part of it. It's still there.
so how will i change it into \[\sec \theta d \theta\]