<s>[INST] How to calculate $ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $ How to calculate 
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $$

I already know one possible way, that is by :
$$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx $$
$$= x-
\int \frac{1}{1+\tan^{6}(x)} dx $$
Then letting $u=\tan(x)$, we must solve
$$\int \frac{1}{(1+u^{6})(1+u^{2})} du $$
We can reduce the denominator and solve it using Partial Fraction technique. This is quite tedious, I wonder if there is a better approach.

Using same approach, for simpler problem, I get
$$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$
 [/INST] An alternative would be to do a load of trig manipulation: 
Starting with $$\frac{s^6}{s^6+c^6}=1-\frac{c^6}{s^6+c^6}$$
$$=1-\frac{\frac 18(1+\cos2x))^3}{ \frac 18(1+\cos2x))^3+\frac 18(1-\cos2x))^3}$$
= a few more lines of algebra = $$=\frac 12+\frac 12\cos 2x\left(\frac{3+\cos^2 2x}{1+3\cos^2 2x}\right)$$
$$=\frac 12+\frac 12\cos 2x\left(\frac{4-\sin^2 2x}{4-3\sin^2 2x}\right)$$
Now integrate, and for the second part us the substitution $$u=\frac{\sqrt{3}}{2}\sin 2x$$
and you end up with $$\frac 12 x+\frac 16\sin x\cos x+\frac{1}{3\sqrt{3}}\operatorname{artanh}(\sqrt{3}\sin x\cos x)+c$$
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