$ \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

  • $\begingroup$ What are the bounds of the integral? You have failed to provide us with vital information! And what is theta? What domain is it operating within. It just appears out of nowhere yet you never said what it is. Is this a complex number integral, vector valued integral? You have yet to make yourself clear. $\endgroup$ – The Great Duck Mar 31 '17 at 0:06
  • $\begingroup$ 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
  • $\begingroup$ @KajHansen I meant what domain does x operate within. They never specified what x was. $\endgroup$ – The Great Duck Mar 31 '17 at 0:09
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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

$$ \int u dv = uv - \int v du $$

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

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


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