1one method i can think of now is put tan(x2) = t
and write tanx = 2tan(x2)1-tan2(x2)
and sinx = 2tanx1+t2
and then use partial fractions..
im sure there'll be a shorter method
1708\hspace{-16}\int\frac{\tan x}{(1-\sin x)}dx$\\\\ Put $x=2t\Leftrightarrow dx=2dt$\\\\ $=2.\int\frac{\tan 2t}{(1-\sin 2t)}d$\\\\ $=2.\int\frac{\sin 2x}{\cos 2x.(1-\sin 2t)}dt$\\\\ $=2.\int\frac{2\sin t.\cos t}{(\cos^2 t-\sin^2 t).(\cos t-\sin t)^2}dt$\\\\ Now Divide $N_{r}$ and $D_{r}$ by $\cos^4 t$\\\\\ $=4.\int\frac{\tan t.\sec^2 t}{(1-\tan^2 t).(1-\tan t)^2}dt=4.\int\frac{\tan t.\sec^2 t}{(1-\tan t)^3.(1+\tan t)}dt$\\\\ Now Put $\tan t=u\Leftrightarrow \sec^2 tdt=du$\\\\ $=4.\int\frac{u}{(1-u)^3.(1+u)}du$\\\\ Now Put $(1-u)=\frac{1}{v}\Leftrightarrow du=\frac{1}{v^2}dv$\\\\ $=4.\int\frac{(1-\frac{1}{v})}{\frac{1}{v^3}.(2-\frac{1}{v})}.\frac{1}{v^2}dv$\\\\ $=4.\int\frac{(v-1).v}{(2v-1)}dv=\int\frac{(2v-1)^2-1}{(2v-1)}dv$\\\\ $=\int(2v-1)dv-\int\frac{1}{(2v-1)}dv$\\\\ $=v^2-v-\frac{1}{2}.\ln\left|(2v-1)\right|+C$\\\\
341\int \frac{\sin x}{\cos x (1- \sin x)} \ dx = \int \frac{\sin x \cos x}{\cos^2 x (1- \sin x)} \ dx
Now let sin x = t
So we have
\int \frac{t}{(1-t)^2 (1+ t)} \ dt = \frac{1}{2} \int \left[ \frac{1}{(1-t)^2} - \frac{1}{(1- t^2)} \right] \ dt
which is easily evaluated.