Simple integral (involving trig)? This seems like a simple problem, but my trig manipulations are leading to a dead end.
Compute: $$\int\frac{\sin^2(x)}{1 - \tan(x)} dx$$
Working thus far:
Replace $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$
Multiply fraction by $\dfrac{\cos(x)}{\cos(x)}$ and then by $\dfrac{\cos(x)+\sin(x)}{\cos(x)+\sin(x)}$ to get:
\begin{align*}
\int\frac{\sin^2(x)\cos(x)(\cos(x) + \sin(x))}{\cos^2(x) - \sin^2(x)} dx &= \int\frac{\sin^2(x)\cos^2(x) + \sin^3(x)\cos(x)}{\cos(2x)} dx \\
&=  
\int\frac{\sin^2(2x)}{4\cos(2x)} dx + \int\frac{\sin^3(x)\cos(x)}{\cos(2x)} dx.
\end{align*}
From there I'm not sure to proceed (in particular on the first integral). The second integral should be doable if I use $\cos(2x) = 1 - 2\sin^2(x)$ and use $ u = \sin(x)$. 

Let $I = \int \dfrac{sin^2x}{1 - tanx} dx$, and $J = \int \dfrac{cos^2x}{1 - tanx} dx$.
Then $I + J = \int \dfrac{1}{1 - tanx} dx$, and $J - I = \int \dfrac{cos^2x - sin^2x}{1 - tanx} dx = \int (cos^2x + sinx\cdot cosx)dx $. By solving these simpler integrals we can then find $I$ ( and $J$).
