$$\int_{\frac{\sqrt{2}}3}^{\frac 23} \frac{dx}{x^5\sqrt{9x^2-1}}$$
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?
Best Answer
Use:
$$\frac{1}{\sec^4(x)} = \cos^4(x) = (\cos^2(x))^2 = (\frac{1 + \cos(2x)}{2})^2 = \frac14(1 + 2\cos(2x) + \cos^2(2x)) = \frac14(1 + 2\cos(2x) + \frac{1 + \cos(4x)}{2}) = etc. $$