Integral of $\frac{-2}{\sqrt{16-x^2}}$ So I am asked to find the anti-derivative of
$$\frac{-2}{\sqrt{16-x^2}}$$
First step I took was to make it easier for me to visualise
$$\int{-2(16-x^2)^{-\frac{1}{2}}}dx$$
I let $u = 16-x^2$
So $\frac{du}{dx} = -2x du$
So substituting in $u$ we have
$$=\int{-2(u)^{-\frac{1}{2}}}dx$$  
Substituting in $\frac{du}{dx}$ we have
$$=\int{\frac{u^{-\frac{1}{2}}}{x}}du$$
Since $x$ is a constant I move it to the front
$$=\frac{1}{x}\int{{u^{-\frac{1}{2}}}}du$$
The integral of $u^{-1/2}du$ is $2u^{1/2}$ +c
So I have
$$=\frac{2\sqrt{u}}{x} + c$$
Substituting $u$ back in I have 
$$=\frac{2\sqrt{16-x^2}}{x}+c$$
I know I can do this by trigonometric substitution however that wouldn't be obvious to me at this stage and therefore would only be used as a last resort. The first thing I would try is simple integration by substitution but I can't see where I have gone wrong. 
 A: We have
$$-2\int\frac{dx}{\sqrt{16-x^2}}=-2\int\frac{d\left(\frac x4\right)}{\sqrt{1-\left(\frac{x}{4}\right)^2}}=-2\arcsin\left(\frac x4\right)+C$$
A: trigonometric substitution isn't that difficult. It just requires practice.
try expressing
$$\frac{-2}{\sqrt{16-x^2}}$$
as
$$\frac{-2}{\sqrt{16(1-(\frac{x}{4})^2)}} = \frac{-2}{4\sqrt{(1-(\frac{x}{4})^2)}}=\frac{-1}{2\sqrt{(1-(\frac{x}{4})^2)}}=-\frac{1}{2\sqrt{(1-\sin^2(\sin^{-1}(\frac{x}{4})))}}$$
so for the integral
$$\int\frac{-2}{\sqrt{16-x^2}}dx = -\frac{1}{2}\int\frac{1}{\sqrt{1-\sin^2(\sin^{-1}(\frac{x}{4}))}}dx$$
If we try to express $\frac{x}{4}$ as $\sin(\sin^{-1}(\frac{x}{4}))$ the substitution becomes obvious.
$$u = \sin^{-1}(\frac{x}{4})\quad\text{or}\quad 4\sin u=x$$
$$4\cos u\,du/dx=1$$
$$4\sqrt{1-\sin^2 u}\,du/dx=1$$
$$dx=4\sqrt{1-\sin^2 u}\,du$$
and our integral becomes
$$-\frac{1}{2}\int\frac{1}{\sqrt{1-\sin^2u}}4\sqrt{1-\sin^2 u}\,du=-2\int du=-2u+C=-2\sin^{-1}(\frac{x}{4})+C$$
Just one final note on integrating this way...suppose that we integrated a different problem with $u=\sin^{-1}x$, and our final answer was $\tan u+C$, we would want to express the $\tan$ in terms of $\sin$ as follows.
$$\tan u + C = \frac{\sin u}{\cos u}+C=\frac{\sin u}{\sqrt{1-\sin^2 u}}+C$$
this allows for the $\sin$s to cancel out with the $\sin^{-1}$s.
$$\frac{\sin u}{\sqrt{1-\sin^2 u}}+C=\frac{\sin(\sin^{-1}x)}{\sqrt{1-\sin^2(\sin^{-1}x)}}+C=\frac{x}{\sqrt{1-x^2}}+C$$
In general, before deciding on the substitution, we express
$$1-x^2 = 1-\sin^2(\sin^{-1}x)=\cos^2(\sin^{-1}x);$$
$$x^2-1 = \sec^2(\sec^{-1}x)-1=\tan^2(\sec^{-1}x);$$
$$x^2+1 = \tan^2(\tan^{-1}x)+1=\sec^2(\tan^{-1}x).$$
Like I said before, it takes practice, but it does make trigonometric more intuitive and less of just an algorithm.
