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Two simple integration

\begin{align}\int \frac{\cos (2 x)}{\cos (x)-\sin (x)} \, dx\end{align}

\begin{align}\int \frac{\cos (2 x)}{\cos ^2(x) \sin ^2(x)} \, dx\end{align}

Make $\cos (x)-\sin (x)$ to product form?

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    $\begingroup$ Just use $\cos(2x) = \cos^2(x) - \sin^2(x)$. Use it raw in the second integral. Factor it in the first integral. $\endgroup$
    – Tunococ
    Sep 17 '13 at 7:20
  • $\begingroup$ @Tunococ Thanks, it's really easy, :). $\endgroup$
    – polynomial
    Sep 17 '13 at 7:23
  • $\begingroup$ Maple can help you. $\endgroup$
    – user64494
    Sep 17 '13 at 7:52
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In fact it is just a piece of cake.

$\int\dfrac{\cos2x}{\cos x-\sin x}dx$

$=\int\dfrac{\cos^2x-\sin^2x}{\cos x-\sin x}dx$

$=\int\dfrac{(\cos x+\sin x)(\cos x-\sin x)}{\cos x-\sin x}dx$

$=\int(\cos x+\sin x)~dx$

$=\sin x-\cos x+C$

$\int\dfrac{\cos2x}{\cos^2x\sin^2x}dx$

$=\int\dfrac{4\cos2x}{\sin^22x}dx$

$=\int\dfrac{2~d(\sin2x)}{\sin^22x}$

$=-2\csc2x+C$

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