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  $\int-5 \sin x \tan x \,dx$

Following your steps, with corrections:
$\int\frac{{\sqrt {25-x^2}}}x\,dx$
I see that it is the form $a-x^2$.
Let $x = 5\sin\theta$
Then $dx=5\cos\theta d\theta$
Then substitute this $x$ and $dx$ value in and 
get $\int\frac {\sqrt {25-(5 \sin\theta)^2}}{5 \sin\theta}\, 5\cos\theta d\theta $
Using the trig identity, simplify the root to get  $\int\frac {\sqrt {25\cos^2\theta}}{5 \sin\theta}\, 5\cos\theta d\theta $
Factor out 5 and cancel, simplified to get  $\int\frac {5|cos\theta|}{ \sin\theta}\, \cos\theta d\theta $
(Note that we're taking the square root of a square here, so we should get the absolute value. Unless you're on a domain where cosine is positive, you can't neglect this.)
This seems to not be where you want to go.
Let's try instead a substitution of $x=5\cos\theta$. Then $dx=-5\sin\theta d\theta$
Thus, we have $-\int\frac{\sqrt{25-(5\cos\theta)^2}}{5\cos\theta}5\sin\theta d\theta$
After simplification, we get $-5\int\frac{|\sin\theta|}{\cos\theta}\sin\theta d\theta$
The simplification you get works if x is such that $\sin\theta$ is always positive, so we can drop the absolute value sign.
