How to properly simplify and trig substitute this integral? $\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then substitute this $x$ value in and 
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, d\theta $
Take the root, simplified to $\int\frac { {5-(5 \sin\theta)}}{5 \sin\theta}\,d\theta $
Factor out 5 and cancel, simplified to $\int\frac { {5(1-( \sin\theta))}}{5 (\sin\theta)}\, d\theta $
Divide by $\sin\theta$ to get $\int\frac { (1-( \sin\theta))}{ (\sin\theta)}\, d\theta $
Simplified to $\int\frac {1}{ \sin\theta}d\,\theta - 1 $
I am stuck here and I can't figure it out. Supposedly the answer simplifies to  $\int-5 \sin x \tan x \,dx$

Let us begin by $x=5t $ and $dx=5\,dt $.
it becomes
$$\int\frac {\sqrt {25 (1-t^2)}}{5t}5\,dt $$
$$=5\int \frac {\sqrt {1-t^2}}{t}\,dt $$
now put $t=\sin (u) $ and $dt=\cos (u)\,du $.
we get
$$\pm 5\int \frac {\cos(u)}{\sin (u)}\cos (u)\,du $$
$$=\pm 5 \int \frac {\cos^2 (u)}{1-\cos^2 (u)}\sin (u)\,du $$
put $v=\cos (u) $ and you can finish.
