<s>[INST] Calculate the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\tan ^4 x}{\cos 2x} dx$ Calculate the integral 
$$\int\limits_{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}}\dfrac{\tan ^4 x}{\cos 2x} dx$$
 [/INST] I evaluate this integral via its Cauchy Principal Value.  To wit, note that
$$\cos{2 x} = \frac{2}{\sec^2{x}}-1$$
so that the integral takes the form
$$\int_{\pi/3}^{\pi/6} dx \sec^2{x} \frac{\tan^4{x}}{1-\tan^2{x}}$$
We can then let $u=\tan{x}$ and substitute to get
$$\int_{1/\sqrt{3}}^{\sqrt{3}} du \frac{u^4}{1-u^2} = \int_{1/\sqrt{3}}^{\sqrt{3}} du \frac{1}{1-u^2} - \int_{1/\sqrt{3}}^{\sqrt{3}} du \: (1+u^2)$$
The first integral on the RHS blows up, but its Cauchy PV is zero by symmetry about the pole at $u=1$.  The other integral is straightforward; the result is
$$\begin{align}PV \int_{\pi/3}^{\pi/6} dx \frac{\tan^4{x}}{\cos{2 x}} &= -\left (\sqrt{3}-\frac{1}{\sqrt{3}} \right) - \frac{1}{3}\left (3\sqrt{3}-\frac{1}{3\sqrt{3}} \right) \\ &= -\frac{44}{9 \sqrt{3}}\end{align}$$
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