difficulty working out equation $\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $ =$\left.2[\frac{1}{2}\sqrt{a^2-x^2}+a^2\arcsin(\frac{x}{a})]\right|_{-a}^{a}$ I encountered the below formula in my text.
I know the author is using integration by substitution and double angle formula
but for some reason, every working on paper that i did is different from the answer below:  
$\int_{-a}^{a}2(\sqrt{a^2-x^2})dx $
=$\left.2[\frac{1}{2}(\sqrt{a^2-x^2}+a^2\arcsin(\frac{x}{a}))]\right|_{-a}^{a}$   
Can anyone help me to expand the above equation.
I can't wrap my head on how above answer is reproduced.
Thanks.  
 A: This can be done with a standard $\sin$ substitution: $(1)\ x=a\sin(\theta)$ followed by a change of variables: $(2)\ \phi=2\theta$.
$$
\begin{align}
\int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\int_{-\pi/2}^{\pi/2}2a\cos(\theta)\,\mathrm{d}a\sin(\theta)\tag{1}\\
&=2a^2\int_{-\pi/2}^{\pi/2}\cos^2(\theta)\,\mathrm{d}\theta\\
&=2a^2\int_{-\pi/2}^{\pi/2}\frac{1+\cos(2\theta)}{2}\,\mathrm{d}\theta\\
&=2a^2\int_{-\pi}^\pi\frac{1+\cos(\phi)}{4}\,\mathrm{d}\phi\tag{2}\\
&=2a^2\left[\frac{\phi+\sin(\phi)}{4}\right]_{\phi=-\pi}^{\phi=\pi}\\
&=\pi a^2
\end{align}
$$
However, the answer you show in the question looks like the answer that comes from an integration by parts: $u=\sqrt{a^2-x^2}$ and $\mathrm{d}v=\mathrm{d}x$ so that $v=x$ and $\mathrm{d}u=-\frac{x\,\mathrm{d}x}{\sqrt{a^2-x^2}}$
$$
\begin{align}
\int2\sqrt{a^2-x^2}\,\mathrm{d}x
&=2x\sqrt{a^2-x^2}+\int\frac{2x^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\\
&=2x\sqrt{a^2-x^2}-\color{#C00000}{\int\frac{2(a^2-x^2)}{\sqrt{a^2-x^2}}\,\mathrm{d}x}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\tag{3}
\end{align}
$$
Adding the integral in red to both sides of $(3)$ and dividing by $2$ yields
$$
\begin{align}
\int2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\frac12\left[2x\sqrt{a^2-x^2}+\int\frac{2a^2}{\sqrt{a^2-x^2}}\,\mathrm{d}x\right]\\
&=\frac12\left[2x\sqrt{a^2-x^2}+2a^2\,\sin^{-1}(x/a)\right]+C\tag{4}
\end{align}
$$
Evaluating $(4)$ at the limits of integration yields
$$
\begin{align}
\int_{-a}^a2\sqrt{a^2-x^2}\,\mathrm{d}x
&=\left[x\sqrt{a^2-x^2}+a^2\,\sin^{-1}(x/a)\right]_{-a}^a\\
&=\pi a^2
\end{align}
$$
Not quite what you got, but of the same form.
A: let $x = a \sin(u)$
then use
$\sin^2(u) + \cos^2(u) = 1$
A: Step-by-step solution from WolframAlpha
$$2\int_{ - a}^a {\sqrt {{a^2} - {x^2}} dx}  = 4\int_0^a {\sqrt {{a^2} - {x^2}} dx} $$
$$\eqalign{
  & x = a\sin t  \cr 
  & dx = a\cos t \cr} $$
$$x\to a\Rightarrow t\to\pi /2\\x\to 0\Rightarrow t\to 0$$
$$\eqalign{
  &  = 4a\int_0^{\pi /2} {a\sqrt {1 - {{\sin }^2}t} \cos tdt}   \cr 
  &  = 4{a^2}\int_0^{\pi /2} {{{\cos }^2}tdt}   \cr 
  &  = 4{a^2}\int_0^{\pi /2} {\frac{{1 + \cos 2t}}{2}dt}   \cr 
  &  = 4{a^2}\left. {\frac{{2t + \sin 2t}}{4}} \right|_0^{\pi /2} = 4{a^2}\frac{\pi }{4} = \pi {a^2} \cr} $$
