Are you by chance referring to the identity

$\displaystyle \frac3{1+\sin^3 x}=\dfrac1{1+\sin x}+\frac1{1-\omega_1\sin x}+\frac1{1-\omega_2\sin x}$

where $\displaystyle \omega_1$, $\displaystyle \omega_2$ are the complex cube roots of $\displaystyle -1$, together with the substitution $\displaystyle t=\tan{\textstyle\frac12}x$ for $\displaystyle \int\dfrac1{1-a\sin x}\mathrm dx$ which can be found when $\displaystyle a^2\neq 1$ using $\displaystyle \int\frac1{1-2at+t^2}\mathrm dt=\frac{\arctan\left(\displaystyle\frac{t-a}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}}$? No, I haven't tried it this way (;->)