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$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$

What I did was trig substitution: $$x=\frac 13 \sec \theta$$ $$dx=\frac 13 \sec \theta \tan \theta \, d\theta$$

Then my integral becomes $$\int_{\frac 13 \sec\frac{\sqrt{2}}{3}}^{\frac 13 \sec\frac 23} \frac{81}{\sec^4 \theta} \, d\theta$$

But I got stuck here. What can I do next?

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    $\begingroup$ $\frac{1}{\sec^4 t} = \cos^4 t $. $\endgroup$ – science Mar 14 '15 at 20:58
  • $\begingroup$ Use the above and cos(2t) = cos^2(t) - 1 or De Moivre's Theorem if you're feeling fancy! $\endgroup$ – user142671 Mar 14 '15 at 21:10
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Use:

$$\frac{1}{\sec^4(x)} = \cos^4(x) = (\cos^2(x))^2 = (\frac{1 + \cos(2x)}{2})^2 = \frac14(1 + 2\cos(2x) + \cos^2(2x)) = \frac14(1 + 2\cos(2x) + \frac{1 + \cos(4x)}{2}) = etc. $$

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