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I am having real trouble evaluating this indefinite integral

$$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$

Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ terms.

Here is what I have already tried:

  1. Seperating denominator as $(1-\sqrt{2}\sin{x}\cos{x})(1-\sqrt{2}\sin{x}\cos{x})$ and applying partial fractions.
  2. Multiplying and dividing by $\sec^4{x}$

Please provide any help. It would be appreciated if you give a hint rather than the whole solution.

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    $\begingroup$ If I was up to it, my first try would be to use $$\sin (x)=\dfrac{2\tan (x/2)}{1+(\tan (x/2))^2}\\ \cos(x)=\dfrac{1-(\tan (x/2))^2}{1+(\tan (x/2))^2}$$ and make the substitution $u=\tan (x/2)$. However, I avoid masochism as much as possible. $\endgroup$ – Git Gud Sep 14 '13 at 11:29
  • $\begingroup$ @GitGud I am scared by the thought, but I'll give it a shot anyways! Any alternative ideas? $\endgroup$ – Cheeku Sep 14 '13 at 11:33
  • $\begingroup$ None other than the standard tries. I'm in hopes that there is an elegant trick that spits out the answer immediately, but since you said WA gave a complicated answer, I don't think such an easy alternative exists. $\endgroup$ – Git Gud Sep 14 '13 at 11:37
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$$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2$$

If we set $\int(\sin x+\cos x)dx=\sin x-\cos x,\sin2x=1-u^2$

$$\implies \int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx=2\int\frac{du}{2-(1-u^2)^2}=2\int\frac{du}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}$$

$$=\frac2{2\sqrt2}\int\frac{(\sqrt2+1-u^2)+(\sqrt2-1+u^2)}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}du$$

$$=\frac1{\sqrt2}\left(\int\frac{du}{\sqrt2-1+u^2}+\int\frac{du}{\sqrt2+1-u^2}\right)$$

Now apply $\displaystyle\int\frac{dy}{a^2+y^2}=\frac1a\arctan \frac ya+K$

and $\displaystyle\int\frac{dy}{a^2-y^2}=\frac1{2a}\ln\big|\frac{a+y}{a-y}\big|+C$

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