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

If $x = 5\sin\theta$ then $dx = 5\cos\theta\, d\theta.$ That was neglected.
You also replaced $\sqrt{25 - (5\sin\theta)^2}$ with $5 - 5\sin\theta.$ That is also not correct. Notice that, for example,
$$
\sqrt{5^2 - 3^3} = \sqrt{25-9} = \sqrt{16} = 4 \ne 5-3, \text{ so } \sqrt{5^2-3^2} \text{ differs from } 5-3.
$$
