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$$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$

I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:

$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2}x = 1 - \frac{1}{2}\cdot\sin^{2}(2x) = \frac{1 + \cos^{2}(2x)}{2}$,

and then using the $\tan(2x) = t$ substitution. But if I do the same with the definite integral, both bounds of the integral become $0$.

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We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$

$\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$

$$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$

Setting $\tan2x=u,$ $$\int\frac{2\sec^22x}{\tan^22x+2}dx=\int\frac{du}{u^2+(\sqrt2)^2}=\frac1{\sqrt2}\arctan\left(\frac u{\sqrt2}\right)+K$$

$$\implies \int\frac{dx}{\cos^4x+\sin^4x}=\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\ \ \ \ (1)$$

Now $\displaystyle\tan2x=0\iff 2x=n\pi\iff x=\frac{n\pi}2$ where $n$ is any integer

Establish that $$\int_0^{2a}f(x)dx=\begin{cases} 2\int_0^af(x)dx &\mbox{if } f(2a-x)=f(x) \\ 0 & \mbox{if } f(2a-x)=-f(x) \end{cases} $$

Setting $2a=2\pi\iff a=\pi$ and $\displaystyle f(x)=\cos^4x+\sin^4x$

$\displaystyle\cos(2\pi-x)=\cos x,\sin(2\pi-x)=-\sin x\implies f(2\pi-x)=f(x)$

$$\implies I=\int_0^{2\pi}\frac{dx}{\cos^4x+\sin^4x}=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}$$

Again setting $\displaystyle2a=\pi\iff a=\frac\pi2$

$\displaystyle\cos(\pi-x)=-\cos x,\sin(\pi-x)=+\sin x\implies f(\pi-x)=f(x)$

$$\implies I=2\int_0^{\pi}\frac{dx}{\cos^4x+\sin^4x}2=2\cdot2\int_0^{\dfrac\pi2}\frac{dx}{\cos^4x+\sin^4x}$$

Finally set $\displaystyle2a=\frac\pi2\iff a=\frac\pi4$

$\displaystyle\implies\cos\left(\frac\pi2-x\right)=\sin x,\sin\left(\frac\pi2-x\right)=\cos x$ $\displaystyle\implies f(x)=f\left(\frac\pi2-x\right)$

$\displaystyle\implies I=4\cdot2\int_0^{\dfrac\pi4}\frac{dx}{\cos^4x+\sin^4x}$

From $\displaystyle(1),I=8\left[\frac1{\sqrt2}\arctan\left(\frac{\tan2x}{\sqrt2}\right)+K\right]_0^{\frac\pi4}=\frac8{\sqrt2}\left(\frac\pi2-0\right)$

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  • $\begingroup$ So, in my case, i have $\int_0^{2\pi} \frac{dx}{\sin^{4}x + cos^{4}x} = 4\cdot \int_0^\pi \frac{dx}{1 + \cos^{2}(2x)}$, and if I break that integral in two parts ($0$ to $\frac{\pi}{2}$ and $\frac{\pi}{2}$ to $\pi$), I still get $0$ as the end result. $\endgroup$ – Proka Apr 15 '14 at 14:49
  • $\begingroup$ @Proka, Please find the edited version $\endgroup$ – lab bhattacharjee Apr 15 '14 at 17:42
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Let $z=e^{i x}$; then $dx = -i dz/z$ and the integral is equal to

$$-i 8 \oint_{|z|=1} dz \frac{z^3}{z^8 + 6 z^4+1}$$

By the residue theorem, the integral is then equal to $i 2 \pi$ times the sum of the residues at each pole inside the unit circle. The residue at each pole $z_k$ is equal to

$$-i 8 \frac{z_k^3}{8 z_k^7 + 24 z_k^3} = -i \frac1{z_k^4+3}$$

Each pole $z_k$ inside the unit circle satisfies $z_k^4+3=2 \sqrt{2}$, so the integral is therefore

$$2 \pi \frac{4}{2 \sqrt{2}} = 2 \sqrt{2} \pi$$

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  • $\begingroup$ That's a nifty way of determining the residues of the poles inside of the unit circle. Very nice. Wolfram Alpha can't determine the residues. $\endgroup$ – Random Variable Apr 15 '14 at 14:26
  • $\begingroup$ @RandomVariable: I have learned the hard way that, many times, it is worth finding an expression for the residues in terms of unknown poles before going ahead and finding those poles. This is an occasion where that strategy pays off big time. $\endgroup$ – Ron Gordon Apr 15 '14 at 14:32
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    $\begingroup$ This does give the correct solution, but it's outside of the scope of the course I'm taking. $\endgroup$ – Proka Apr 15 '14 at 14:38
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    $\begingroup$ @RonGordon: I thought so, I upvoted your solution for visibility, I just wanted to let you know why I can't choose it as my accepted answer. Thank you nonetheless. $\endgroup$ – Proka Apr 15 '14 at 14:52
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    $\begingroup$ @N3buchadnezzar: $$z^8+6 z^4+1=0 \implies z^4=-3 \pm 2 \sqrt{2}$$ The only roots in the circle are the $+$ roots. $\endgroup$ – Ron Gordon Apr 15 '14 at 16:05
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Note that $\tan 2x$ is discontinuous at $\frac{\pi}4$ and some other values which can be easily found.

You have to break the integral from $0$ to $\frac{\pi}4$ and so on.

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  • $\begingroup$ You might be on track here, but it's not discontinous at $\pi$, it's discontinous at $\frac{\pi}{4}$, because $tan\frac{\pi}{2} = \infty$ $\endgroup$ – Proka Apr 15 '14 at 13:14
  • $\begingroup$ @Proka Oh yeah, thanks. Edited. $\endgroup$ – evil999man Apr 15 '14 at 13:16
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    $\begingroup$ @Proka $\tan \pi/2$ does not equal $\infty$ Its just undefined. Not even the limit exists. It will be clear if you see its graph.(google tan(x)) $\endgroup$ – evil999man Apr 15 '14 at 18:27

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