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!
 A: 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.
A: Here's one more way to do it. Using the substitution $x=\tan{\theta}$,
$$\begin{align}
\int_{\pi/6}^{\pi/4}\frac{4\,\mathrm{d}\theta}{\sin^2{\left(\theta\right)}\cos^2{\left(\theta\right)}}
&=4\int_{1/\sqrt{3}}^{1}\frac{(x^2+1)^2}{x^2}\cdot\frac{\mathrm{d}x}{x^2+1}\\
&=4\int_{1/\sqrt{3}}^{1}\frac{x^2+1}{x^2}\mathrm{d}x\\
&=4\int_{1/\sqrt{3}}^{1}\left(1+\frac{1}{x^2}\right)\mathrm{d}x\\
&=4\left(x-\frac{1}{x}\right)_{1/\sqrt{3}}^{1}\\
&=\frac{8}{\sqrt{3}}.
\end{align}$$
A: Another way:
Use $\sin^2x + \cos^2 x = 1$
$\displaystyle \int dx\frac{\sin^2x + \cos^2x}{\sin^2x \cos^2x} = \int dx\sec^2x + \csc^2 x = \tan x - \cot x +C$
A: Write the integral as

$$ \int_{\pi/6}^{\pi/4} \sec^2(x) \csc^2(x) dx $$

then use integration by parts with $u=\csc^2(x)$ gives

$$ 2-\frac{4}{\sqrt {3}}\,+2\,\int _{1/6\,\pi }^{1/4\,\pi }\! \csc^2(x){dx}.$$

I think you can finish it.
A: Note that $\sin(2x) = 2\sin(x)\cos(x)$
Thus your integral can be rewritten as
$$ \int \frac{16}{\sin(2x)^2} dx $$
course we can make the change $2x = u$ to obtain
$$ 8\int \frac{1}{\sin(u)^2} du $$
Which is just 
$$8 \cot(2x) + C$$
A: Use the fact that $\sin2x=2\sin x\cos x$, then, after first letting $t=2x$, employ the Weierstrass substitution.
