$\int \frac{\sin^3x}{\sin^3x + \cos^3x)}$?

Well, I'm still not seeing any nice ways of doing it. I do see at least one way of proceeding though. First, divide top and bottom by $\cos^3x$

$$\int\frac{\tan^3xdx}{1+\tan^3x}$$

Now make the substitution

$$x=\tan^{-1}u,dx=\frac{du}{1+u^2}$$

$$\int\frac{u^3du}{(1+u^2)(1+u)(1-u+u^2)}$$

at which point it can be solved by partial fractions.

Put $$I = \displaystyle\int{\dfrac{\sin^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x , \quad J = \displaystyle\int{\dfrac{\cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x.$$ We have $$I + J = \displaystyle\int{\mathrm{d}x} = x+ C.$$ and \begin{equation*} I - J = \displaystyle\int{\dfrac{\sin^3 x - \cos^3 x}{\cos^3 x +\sin^3 x }}\mathrm{d}x = \displaystyle\int{\dfrac{(\sin x - \cos x)(1 + \sin x \cdot \cos x)}{(\sin x + \cos x)(1 - \sin x \cdot \cos x)}\mathrm{d}x} \end{equation*} Put $t = \sin x + \cos x$, then $\mathrm{d}t = -(\sin x - \cos x) \mathrm{d}x$ and $ \sin x \cdot \cos x = \dfrac{ t^2-1}{2}.$ We get \begin{equation*} I - J = \displaystyle\int{\dfrac{t^2 + 1}{t(t^2-3)}\mathrm{d}t} = \dfrac{2}{3}\ln{(t^2-3)}-\dfrac{1}{3}\ln t + C'. \end{equation*} and then \begin{equation*} I - J = \dfrac{2}{3}\ln{((\sin x + \cos x)^2-3)}-\dfrac{1}{3}\ln (\sin x + \cos x) + C'. \end{equation*} From $I + J $ and $I - J$, we can calculate $I$.

If all else fails, the Weierstrass substitution will do it.