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
$$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$
 A: 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 $$\frac 12 x+\frac 16\sin x\cos x+\frac{1}{3\sqrt{3}}\operatorname{artanh}(\sqrt{3}\sin x\cos x)+c$$
A: Let us take:
$$I=\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$
then
$$I=\int \frac{-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
giving$$2I=\int \frac{\sin^{6}(x)-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$
This can be written as (using identities like $a^3-b^3$ and $a^3+b^3$)
$$2I= \int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x)(1+\sqrt3\sin(x)\cos(x))}dx+x$$
$$
2I=\frac{1}{2}\left(\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))} 
+\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x))}\right)+x
$$
Evaluating the integrals separately by using $u=1+\sqrt3\sin(x)\cos(x)$
for first one gives
$$\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))}=\frac{1}{\sqrt3}\int\frac{(\sin(x)\cos(x)-1)(\sin(x)\cos(x)+1)}{u}du$$
Now use $\sin(x)\cos(x)=\frac{u-1}{\sqrt3}$
which will evaluate the integral as $\frac{u^2}{6\sqrt3}-\frac{2u}{3\sqrt3}-\frac{2\ln(u)}{3\sqrt3}$...Similar approach for other with $v=1-\sqrt3\sin(x)\cos(x)$
The final value is 
$$I=\frac{x}{2}-\frac{\sin(x)\cos(x)}{6}+\frac{\ln(1-\sqrt3\sin(x)\cos(x))}{6\sqrt3}-\frac{\ln(1+\sqrt3\sin(x)\cos(x))}{6\sqrt3}+C$$
A: The partial fractions are not difficult to find if you use $z$ to denote $u^2$. This makes no difference as we are just using it to find partial fraction.
$$\frac{1}{(1+u^6)(1+u^2)} = \frac{1}{(1+z)^2(z^2-z+1)} \\= \frac{A}{1+z}+\frac{B}{(1+z)^2} + \frac{Cz+D}{z^2-z+1}$$
Finding $A$ is simple using heaviside cover rule, we get $B= \frac{1}{3}$. Now $A,C,D$ follow from three equations, which I leave as a task for you. 
Upon finding the coefficients, again denote $z$ as $u^2$ and integrate. 
