Integrate $\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$ Integrate $$\int_0^{2 \pi} \frac{2}{\cos^{6}(x) + \sin^{6}(x)} dx$$
I have tried to do Tangent substitution, but there is huge power.
What method is better to use here?
 A: I suspect the OP did not intend for this, but let's go with the residue theorem.  Even if we make no simplification to the denominator, things end up pretty simple.  Note that, when $z=e^{i x}$,
$$\cos^6{x}+\sin^6{x} = \left (\frac{z+z^{-1}}{2} \right )^6 - \left (\frac{z-z^{-1}}{2} \right )^6 = \frac1{16} \left (3 z^4+10+3 z^{-4} \right )$$
As usual, sub $z=e^{i x}$ in the integral; then the integral is equal to
$$-i 16 \oint_{|z|=1} \frac{dz}{z} \frac2{3 z^4+10+3 z^{-4}} = -i 32 \oint_{|z|=1} dz \frac{z^3}{3 z^8+10 z^4+3}$$
The poles of the denominator occur inside the unit circle where $z_k^4=-1/3$.  By the residue theorem, the integral is equal to $i 2 \pi$ times the sum of the residues at these poles inside the unit circle, or
$$64 \pi \sum_{k=1}^4 \frac{z_k^3}{24 z_k^7+40 z_k^3} = 8 \pi \sum_{k=1}^4 \frac1{3 z_k^4+5} = 8 \pi (4) \left (\frac14 \right ) $$
Thus,

$$\int_0^{2 \pi} dx \frac{2}{\cos^6{x}+\sin^6{x}} = 8 \pi$$

A: HINT:
$$\cos^6x+\sin^6x=(\cos^2x)^3+(\sin^2x)^3=(\cos^2x+\sin^2x)^3-3\cos^2x\sin^2x(\cos^2x+\sin^2x)$$
$$=\dfrac{4-3\sin^22x}4=\dfrac{4(1+\cot^22x)-3}{4\csc^22x}$$
Set $\cot2x=u$
A: $$
\begin{aligned}
\int_0^{2 \pi} \frac{2}{\cos^6 x + \sin^6 x} dx
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^6 x + \sin^6 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{(\cos^2 x  + \sin^2 x)(\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x)} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^4 x-\cos^2 x\sin^2 x+\sin^4 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{1}{\cos^4 x}\frac{1}{1-\tan^2x+\tan^4 x} dx\\
&=8\int_0^{\frac{\pi}{2}} \frac{dx}{\cos^2 x}\frac{1}{\cos^2 x}\frac{1}{1-\tan^2x+\tan^4 x}\\
&=8\int_0^{\frac{\pi}{2}} d(\tan x)\left(1+\tan^2 x\right)\frac{1}{1-\tan^2x +\tan^4 x}\\
&=8\int_0^{+\infty} \frac{(1+t^2)}{1-t^2+t^4} dt\\
&=8\int_0^{+\infty} \left(\frac{1}{t^2}+1\right)\frac{1}{\frac{1}{t^2}-1+t^2} dt\\
&=8\int_0^{+\infty} \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+1}\\
&=8\int_{-\infty}^{+\infty} \frac{ds}{s^2+1}\\
&=8\operatorname{arctan} s|_{-\infty}^{+\infty}\\
&=8\pi 
\end{aligned}
$$
