TITLE: Difficult integral: $\sin^3\theta / (\sin^3\theta - \cos^3\theta)$ 
QUESTION [7 upvotes]: The question is, find the integral to the function:  
$\sin^3\theta / (\sin^3\theta - \cos^3\theta)$  
The only thing I could think of was to factor the denominator. But then I couldn't make any further progress.

REPLY [4 votes]: We take advantage of the symmetry, indeed expand on it.   Let 
$$I=\int \frac{\sin^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta} \qquad\text{and}\qquad J=\int \frac{\cos^3\theta\,d\theta}{\sin^3\theta-\cos^3\theta}.$$
Note that
$$\frac{\sin^3\theta}{\sin^3\theta-\cos^3\theta}=1+  \frac{\cos^3\theta}{\sin^3\theta-\cos^3\theta},$$
and therefore 
$$I-J=\theta.$$
If we can find $I+J$ we will be finished.
So we want to find
$$\int\frac{\sin^3\theta+\cos^3\theta}{\sin^3\theta-\cos^3\theta}\,d\theta=
\int\frac{(\sin\theta+\cos\theta)(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta)}{(\sin\theta-\cos\theta)(\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta) }\,d\theta.$$
Let $u=\sin\theta-\cos\theta$. Then $du=(\cos\theta+\sin\theta)\,d\theta$. Also, 
$u^2=1-2\sin\theta\cos\theta$. From this we find that $\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta=\frac{1+u^2}{2}$ and $\sin^2\theta+\cos^2\theta+\sin\theta\cos\theta=\frac{3-u^2}{2}$.  Thus 
$$I+J=\int\frac{1+u^2}{u(3-u^2)}\,du.$$
We do a partial partial fraction decomposition:
$$\frac{1+u^2}{u(3-u^2)}=\frac{1}{3}\left(\frac{1}{u}+\frac{4u}{3-u^2}\right).$$
Integrate:  $I+J=(1/3)\ln\left(\dfrac{|u|}{(3-u^2)^2}\right).$