How do I solve $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\frac{4\,dx}{\sin^2(x)\cos^2(x)}$? Alright so I have $$\int_{\pi/6}^{\pi/4}\frac{4\,dx}{\sin^2(x)\cos^2(x)}.$$ And I am not completely sure on how to tackle this problem. All I have done thus far is $$4\int_{\pi/6}^{\pi/4}\frac{1}{\sin^2(x)}\frac{1}{\cos^2(x)}dx$$ and I don't know how to approach this problem. Help would be greatly appreciated, thanks in advance!

Recall from trigonometry that $\sin(2x)=2\sin x\cos x$ (so $\sin x\cos x = \frac 1 2\sin(2x)$).
$$
\int \frac{4\,dx}{\sin^2 x\cos^2 x} = \int\frac{4\,dx}{\frac14\sin^2(2x)}.
$$
That reduces to $\displaystyle\int\csc^2 u\,du$ which is tabulated in every textbook.
