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I have a integral which seems difficult to me. Any help would be appreciated.

Find $$\int \frac{\cos5x + 5\cos3x +10\cos x }{\cos6x+ 6\cos4x + 15\cos2x +10}\mathrm dx$$

Also I wound like to know your thought process to solve integrals like these.

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  • $\begingroup$ Express everything with $\cos(x)$ and $\sin(x)$. Then put $t=tg(\frac{x}{2})$. $\endgroup$ – pointer Aug 21 '14 at 12:09
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    $\begingroup$ There might be a clever way with trig identities. But the only insight that immediately springs to mind is to replace $\cos nx$ with $\frac 12(e^{inx} + e^{-inx})$, then make the sub $u = e^{ix}$ and hope to god the algebra (factorisation of the denominator, partial fractions) is simple later on. $\endgroup$ – Deepak Aug 21 '14 at 12:09
  • $\begingroup$ D'Moivers theorem can be used to find $\cos nx$ and $\sin nx$ in terms of $\cos $ and $\sin $ $\endgroup$ – Bumblebee Aug 21 '14 at 12:25
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If you expand the cosines of multiple angles as functions of $\cos(x)$, you should arrive to $$\cos5x + 5\cos3x +10\cos x=16 \cos ^5 x$$ $$\cos6x+ 6\cos4x + 15\cos2x +10=32 \cos ^6 x$$ which means that $$\int \frac{\cos5x + 5\cos3x +10\cos x }{\cos6x+ 6\cos4x + 15\cos2x +10} \mathrm dx=\frac{1}{2}\int\sec x ~\mathrm dx$$ Now, use Weierstrass substitution.

I am sure that you can take from here.

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  • $\begingroup$ Very neat. Could you just "see" that the powers of cosines would simplify like that or did you take a "leap of faith" and actually do the expansion? $\endgroup$ – Deepak Aug 21 '14 at 12:16
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    $\begingroup$ The only thing I "saw" is that only odd angles appear in numerator and even angles appear in denominator. So, I was (almost) sure that all of that had to simplify a lot. $\endgroup$ – Claude Leibovici Aug 21 '14 at 12:18
  • $\begingroup$ Very good insight. The coefficients are chosen carefully so that they will cancel out. If I chose different coefficients, I don't think the problem would be as trivial. $\endgroup$ – Deepak Aug 21 '14 at 12:22
  • $\begingroup$ @Deepak. You are perfectly right ! I am sure that this is the way to build this kind of problem : start with the expansion of $\cos^nx$ as a function of multiple angles and ask the student to do the reverse ! Cheers :_) $\endgroup$ – Claude Leibovici Aug 21 '14 at 12:25
  • $\begingroup$ Well, I had an inkling that something like this was going on because the coefficients looked like binomial coefficients (Pascal's triangle), but (unlike you), I didn't have the "intestinal fortitude" (i.e. guts) to actually attempt the expansion in case it went nowhere! :) Cheers. $\endgroup$ – Deepak Aug 21 '14 at 12:27

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