Can $\int_0^{2\pi} \frac{dx}{\sin^6x+\cos^6x}$ be solved using $\cot x = u$ as substitution? My first guess is it can't, since when I substitute the boundaries, I end up with $\cot2\pi$ and $\cot0$. Nevertheless I tried substituting pretending it is indefinite integral, but I can't get anything meaningful.
 A: Using
$$ \int_0^\infty\frac{x^n}{1+x^m}dx= \frac{\pi  \csc \left(\frac{\pi 
   (n+1)}{m}\right)}{m} $$
(see Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only)
it is easy to see
\begin{eqnarray}
\int_0^{2\pi} \frac{dx}{\sin^6x+\cos^6x}&=&4\int_0^{\frac{\pi}{2}} \frac{dx}{\sin^6x+\cos^6x}\\
&=&4\int_0^{\frac{\pi}{2}} \frac{\csc^6x\,dx}{1+\cot^6x}\\
&=&4\int_0^{\frac{\pi}{2}} \frac{\csc^4x}{1+\cot^6x}\csc^2x\,dx\\
&=&4\int_0^{\frac{\pi}{2}} \frac{(1+\cot^2x)^2}{1+\cot^6x}\csc^2x\,dx\\
&=&4\int_0^\infty\frac{(1+u^2)^2}{1+u^6}\,du\\
&=&\cdots
\end{eqnarray}
A: We have $$\int_{0}^{2\pi}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}dx=4\int_{0}^{\pi/2}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}\cdot\frac{\cos^{4}\left(x\right)}{\cos^{4}\left(x\right)}dx\tag{1}$$ $$=4\int_{0}^{\pi/2}\frac{\left(\tan^{2}\left(x\right)+1\right)\sec^{2}\left(x\right)}{\tan^{4}\left(x\right)-\tan^{2}\left(x\right)+1}dx\overset{\tan\left(x\right)=u}{=}4\int_{0}^{\infty}\frac{u^{2}+1}{u^{4}-u^{2}+1}du
 $$ and now observe that $$\int_{0}^{\infty}\frac{u^{2}+1}{u^{4}-u^{2}+1}du=\int_{0}^{\infty}\frac{u^{2}+1}{\left(u^{2}-1\right)^{2}+u^{2}}du
 $$ $$=\int_{0}^{\infty}\frac{\frac{u^{2}+1}{u^{2}}}{\left(\frac{u^{2}-1}{u}\right)^{2}+1}du\overset{\frac{u^{2}-1}{u}=v}{=}\int_{-\infty}^{\infty}\frac{1}{1+v^{2}}dv=\pi
 $$ hence 

$$\int_{0}^{2\pi}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}dx=4\pi.$$

Note: In $(1)$ we can multiply by $\frac{\sin^{4}\left(x\right)}{\sin^{4}\left(x\right)}$ instead of $\frac{\cos^{4}\left(x\right)}{\cos^{4}\left(x\right)}$ and get $$4\int_{0}^{\pi/2}\frac{\left(\cot^{2}\left(x\right)+1\right)\csc^{2}\left(x\right)}{\cot^{4}\left(x\right)-\cot^{2}\left(x\right)+1}dx$$ and now we can take $\cot(x)=u$. The rest is the same.
A: Hint
$$\sin^6x+\cos^6x=1-\frac{3}{4}\sin^2 2x$$
A: If we consider the problem of the antiderivative $$I=\int \frac{dx}{\sin^6(x)+\cos^6(x)}$$ and use $$\cot(x)=u\implies x=\cot ^{-1}(u)\implies dx=-\frac{du}{u^2+1}$$ after some minor simplifications $$I=-\int\frac{u^2+1}{u^4-u^2+1}\,du=\tan ^{-1}\left(\frac{u}{u^2-1}\right)$$ (just as Marco Cantarini answered).
For the definite integral, use Michael Hardy's comment.
