Prove that

$$ \int \frac1{5S_{8} - 9S_{10} + 7S_{12} - 2S_{14}} {\rm d}x = 2x - \arctan \left( \frac{\tan2x}{2 + \tan^22x} \right) + C$$

where $S_n = \sin^n(x) + \cos^n(x)$.

Even differentiating the right doesn't end up with anything close to that monster integrand.

$$ \frac{{\rm d}}{{\rm d}x} \left( 2x - \arctan \left( \frac{\tan2x}{2 + \tan^22x} \right) + C \right) \\ = 2 + \frac{(\tan^22x+1)(2\tan^22x-4)}{\tan^42x+5\tan^22x+4} \\ = \frac{ 4 } {\sin^42x+ 5\sin^22x\cos^22x + 4\cos^42x} \\ = \frac1{\sin^8x + \sin^2x\cos^6x +\cos^2x\sin^6x + \cos^8x} $$

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    $\begingroup$ The denominator in your final expression is $S_6$ $\endgroup$ – Empy2 Dec 13 '18 at 12:14
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    $\begingroup$ $S_0=2,S_2=1$ $$S_{m+2}=\sin^mx(1-\cos^2x)+\cos^mx(1-\sin^2x)=S_m-\sin^2x\cos^2xS_{m-2}$$ $\endgroup$ – lab bhattacharjee Dec 13 '18 at 14:38

Let $c=\cos2x$. Then applying the double angle identities, $$\sin^2x=\frac{1-\cos2x}2,\quad\cos^2x=\frac{1+\cos2x}2,$$ to each occurrence of $\sin^nx$ and $\cos^nx$, we get $$\begin{cases} S_8=2^{-3}(1+6c^2+c^4)\\ S_{10}=2^{-4}(1+10c^2+5c^4)\\ S_{12}=2^{-5}(1+15c^2+15c^4+c^6)\\ S_{14}=2^{-6}(1+21c^2+35c^4+7c^6) \end{cases}$$ Combining these expressions to match the denominator of the integrand, we find that it reduces to $$5S_8-9S_{10}+7S_{12}-2S_{14}=2^{-2}(1+3c^2)$$ so that $$\int\frac{\mathrm dx}{5S_8-9S_{10}+7S_{12}-2S_{14}}=\int\frac4{1+3\cos^22x}\,\mathrm dx=\int\frac8{5+3\cos4x}\,\mathrm dx$$ The antiderivative is $-\arctan(2\cot2x)+C$, which I don't think should be too hard to reconcile with the given solution.

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    $\begingroup$ The pure elegance and beauty of this answer is incredible. I commend you with a meaningless upvote. $\endgroup$ – clathratus Dec 14 '18 at 16:20

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