I'll restate the accepted answer in different notation, which is easier for me to parse: let
$$u=\sqrt{1-x^2}, \quad dv=dx$$
so that
$$du=\frac{-x}{\sqrt{1-x^2}}dx,\quad v=x$$
For brevity, write $I=\int \sqrt{1-x^2}\, dx$. Using $\int u\,dv = uv-\int v\,du$, obtain
$$
I = x\sqrt{1-x^2} - \int \frac{-x^2}{\sqrt{1-x^2}} \,dx
$$
The last integral does not look simpler than $I$ itself, but it can be related back to it:
$$
\int \frac{-x^2}{\sqrt{1-x^2}} \,dx = \int \frac{1-x^2}{\sqrt{1-x^2}} \,dx - \int \frac{1 }{\sqrt{1-x^2}} \,dx = I - \sin^{-1}x
$$
So,
$$I = x\sqrt{1-x^2} - (I-\sin^{-1}x)$$
and solving for $I$ yields
$$\int \sqrt{1-x^2}\, dx = \frac12 x\sqrt{1-x^2} + \frac12 \sin^{-1}x + C$$
For completeness and comparison, I'll add the conventional solution using $x=\sin t$ substitution. Here $dx=\cos t\,dt$, so
$$
\int\sqrt{1-x^2}\,dx = \int \cos^2 t\,dt =\int \left(\frac12+\frac{\cos 2t}{2}\right)\,dt = \frac{t}{2}+\frac{\sin 2t}{4}+C $$
To return to $x$, note that $t=\sin^{-1}x$ and $\sin 2t = 2\sin t\cos t = 2x\sqrt{1-x^2}$.
First, expand the integral using integration by parts until it reappears:
$$\begin{align*}
\int e^{2t} \sin{t} \ dt & ~=~ \frac{1}{2} e^{2t} \sin{t} - \frac{1}{2} \int e^{2t} \cos{t} \ dt \\
& ~=~ \frac{1}{2} e^{2t}\sin{t} - \frac{1}{4} e^{2t}\cos{t} - \frac{1}{4} \int e^{2t} \sin{t} \ dt
\end{align*}$$
Now combine like terms and multiply by $4 / 5$ to obtain:
$$\begin{align*}
\int e^{2t} \sin{t} \ dt & ~=~ \frac{2}{5} e^{2t}\sin{t} - \frac{1}{5} e^{2t} \cos{t} \\
\end{align*}$$
Best Answer
$e^x \sinh(x) = e^x \frac{e^x - e^{-x}}{2} = \frac{1}{2} \left( e^{2x} -1 \right)$. Why would you want to use integration by parts?