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So I am trying to integrate this problem $\int\frac{\cos(8x)}{\cos(4x)+\sin(4x)}dx$, and my professor went over it in class and went from $\int\frac{\cos(8x)}{\cos(4x)+\sin(4x)}dx \rightarrow \int\cos(4x)-\sin(4x)dx$ and I do not understand how he went about simplifying the integral. I've looked and all the trig identities we've gone over so far in our book and in class and I am not seeing how he got that. Thanks for all the help in advance.

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    $\begingroup$ nice question +1 $\endgroup$ – imranfat Aug 31 '14 at 5:39
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Hint: $\cos(8x)=\cos(2 \cdot 4x)=\cos^2(4x)-\sin^2(4x)=(\cos(4x)+\sin(4x))(\cos(4x)-\sin(4x))$

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Note that $$\cos (2x) = {\cos ^2}x - {\sin ^2}x$$so that $$\eqalign{ & {{\cos (8x)} \over {\cos (4x) + \sin (4x)}} = \cr & {{(\cos (4x) - \sin (4x))(\cos (4x) + \sin (4x))} \over {\cos (4x) + \sin (4x)}} = \cr & \cos (4x) - \sin (4x) \cr} $$

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