Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? 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.
$\int\frac{x^5}{(36x^2+1)^{3/2}}dx=\int\frac{x^5}{((6x)^2+1)^{3/2}}dx$
$x=\frac{1}{6}\tan\theta, dx=\frac{1}{6}\sec^2(\theta) d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\tan^2(\theta)+1)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{(\sec^2\theta)^{3/2}}*\sec^2(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\tan^5(\theta)}{\sec(\theta)}d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^5(\theta)}*\cos(\theta)d\theta$
$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$

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 $u = \cos\theta$ 
