Well, probably not so tough. I haven't done much integrating in a while and I ran into an integral of (sint*tant) in my homework. I couldn't rearrange it in any helpful way, so I'm hoping there is an identity for it. Thanks.
Or you can apply by parts:
$\displaystyle u=\tan x \to du=\sec^2x$ $\displaystyle dv=\sin x \to v=-\cos x$
$\displaystyle \int \sin x\tan x=-\cos x\tan x+\int \cos x\sec^2 x\;dx$
$\displaystyle \sec^2x\cos x = \frac{\cos x}{\cos^2x}=\sec x$
Therefore, $\displaystyle \int \cos x\sec^2 x\;dx=\int \sec x \;dx=\ln|\sec x+\tan x|+C$
Thus, $\displaystyle \int \sin x\tan x\;dx=\ln|\sec x+ \tan x|-\sin x +C$