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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!

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    $\begingroup$ Have you considered the double-angle formulas? $\endgroup$ – abiessu Aug 6 '14 at 20:32
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    $\begingroup$ $$\frac{1}{\sin^2x \cdot \cos^2 x} = \frac{\cos^2x + \sin^2x}{\sin^2x \cdot \cos^2 x}$$ $\endgroup$ – Stephen Dedalus Aug 6 '14 at 20:41
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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.

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Here's one more way to do it. Using the substitution $x=\tan{\theta}$,

$$\begin{align} \int_{\pi/6}^{\pi/4}\frac{4\,\mathrm{d}\theta}{\sin^2{\left(\theta\right)}\cos^2{\left(\theta\right)}} &=4\int_{1/\sqrt{3}}^{1}\frac{(x^2+1)^2}{x^2}\cdot\frac{\mathrm{d}x}{x^2+1}\\ &=4\int_{1/\sqrt{3}}^{1}\frac{x^2+1}{x^2}\mathrm{d}x\\ &=4\int_{1/\sqrt{3}}^{1}\left(1+\frac{1}{x^2}\right)\mathrm{d}x\\ &=4\left(x-\frac{1}{x}\right)_{1/\sqrt{3}}^{1}\\ &=\frac{8}{\sqrt{3}}. \end{align}$$

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Another way:

Use $\sin^2x + \cos^2 x = 1$

$\displaystyle \int dx\frac{\sin^2x + \cos^2x}{\sin^2x \cos^2x} = \int dx\sec^2x + \csc^2 x = \tan x - \cot x +C$

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    $\begingroup$ This method would likely be the MVP for high school students, and I also like the use of the Pythagorean identity for how it hints at sneaky ways to solve similar but harder integrals. +1 $\endgroup$ – David H Aug 6 '14 at 22:23
  • $\begingroup$ @DavidH Yes, it's a pretty clever solution, though it is probably a lot more complex / difficult to see than just substituting $\sin 2x = 2 \sin x \cos x$ $\endgroup$ – MCT Aug 7 '14 at 4:37
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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

$$8 \cot(2x) + C$$

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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.

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    $\begingroup$ Perhaps it is $\sec^2(x)\csc^2(x)$? Where does the extra $\cos^2(x)$ come from? $\endgroup$ – abiessu Aug 6 '14 at 20:49
  • $\begingroup$ @abiessu: Yes you are right. Thanks for the comment. $\endgroup$ – Mhenni Benghorbal Aug 6 '14 at 21:57
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Use the fact that $\sin2x=2\sin x\cos x$, then, after first letting $t=2x$, employ the Weierstrass substitution.

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