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$ \displaystyle\int \frac{x^{3}}{\sqrt{x^{2} + 9}}dx$

doing a trig sub:

$x = 3tan(\theta)$

$dx = 3sec^{2}(\theta)d\theta$

$\displaystyle\int \frac{(3tan(\theta))^{3}}{\sqrt{3tan(\theta)^{2} + 9}}d\theta$

$\displaystyle\int \frac{ ( 27tan^{3}(\theta) } { \sqrt{ 3tan(\theta)^{2} + 9 } }d\theta$

$27\displaystyle\int \frac{tan^{3}(\theta)*sec^{2}(\theta)}{\sqrt{sec^{2}(\theta)}}d\theta$

$27\displaystyle\int tan(\theta)*tan(\theta)*sec(\theta)*d(\theta)$

Got stuck here and think I am doing this wrong. Please help Thank you

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  • $\begingroup$ Your approach looks fine. Keep in mind the pythagorean identity for tan and sec, and this is a solvable integral. Not easy by any means, but solvable. $\endgroup$ – Kaynex Mar 31 '17 at 0:07
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    $\begingroup$ There don't need to be bounds @TheGreatDuck...it's indefinite. Careful to replace the $dx$ with $3\sec^2(\theta)d \theta$ in the first integral after the substitution. $\endgroup$ – Kaj Hansen Mar 31 '17 at 0:07
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    $\begingroup$ Here are a few errors: $x^2=9\tan^2(\theta)$ (but you have another error later, so it doesn't matter), and somehow, you lose a $\tan(\theta)$ in the last step. Now, in the last step, you can replace $\tan^2(\theta)$ again and hopefully recognize what is in front of you. Or you might consider the substitution $u=x^2+9$ instead for your integral. $\endgroup$ – martin.koeberl Mar 31 '17 at 0:13
  • $\begingroup$ Also, you might consider using \tan for $\tan$ and \sec for $\sec$ and \cdot for $\cdot$ (multiplication). $\endgroup$ – martin.koeberl Mar 31 '17 at 0:18
  • $\begingroup$ Thanks! Didn't see that I lost that tan so then now I can do a trig integral: tan^{2}(x)*tan(x)*sec(x)*dx and that to (sec^{2} - 1)*tan(x)*sec(x)dx and then u-sub that (u^2 - 1) take the integral 27 [ sec^{3}(x) / 3 - sec(x) ] creating a triangle I get: 1/9( √(9+x^2) / 3)^3 ) - 27( √(9+x^2) / 3 ) + C .... Is this right or wrong? $\endgroup$ – yre Mar 31 '17 at 0:29
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$$ \int u dv = uv - \int v du $$

$$ u = x^{2}, \qquad dv = \frac{x }{\sqrt{x^{2}+9}}dx $$ $$ du = 2x dx, \qquad v = \sqrt{x^{2}+9} $$


$$ \int \frac{x^{3}}{\sqrt{x^{2}+9}}dx = \frac{1}{3} \left(x^2-18\right) \sqrt{x^2+9} $$

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