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I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've worked these problems before, I usually end up performing a u-substitution after the trig sub, but I'm not sure where to do that or even if this is the right strategy for this integral.

$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$

$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$

$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$

$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$

$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$

$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$

$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$

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You are almost there.

$$\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta=\int\frac{\sin(\theta)(1-\cos^2(\theta))^2}{\cos^4(\theta)}d\theta=-\int\frac{1-2u^2+u^4}{u^4}du\\ =-\int u^{-4}-2u^2+1 du$$

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  • $\begingroup$ Yes, there is a way! Shouldn't that last form $ - \int u^{-4} - \frac{2}{u^2} - 1\ du $, since $\frac{d}{dx}\cos x = - \sin x$ and $\frac{2u^2}{u^4}=\frac{2}{u^2}$? $\endgroup$ – kas Dec 9 '16 at 3:36
  • $\begingroup$ @kas Yes, fixed $\endgroup$ – N. S. Dec 9 '16 at 5:15
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Hint: Let $u=36x^2+1$, then $du=72xdx$ and thus $$\frac{x^5}{(36x^2+1)^{3/2}}dx=\frac{(u-1)^2}{36^2u^{3/2}}\frac{du}{72}.$$ Now we get something easier to handle.

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  • $\begingroup$ Yeah, that's how Mathematica solves it, but is there no solution by trig sub? $\endgroup$ – kas Dec 9 '16 at 3:08
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Hint: Change $$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$$ to $$\frac{1}{6^6}\int\sin(\theta)\tan^4(\theta)d\theta$$ then go to $$\frac{1}{6^6}\int\sin(\theta)(\sec^2(\theta) - 1)^2d\theta$$ then use a u substitution of $u = \cos\theta$

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