Finding $\int \sqrt{3-2x-x^2}dx$ $\int \sqrt{3-2x-x^2}\,dx$
First I did:
$$\begin{align}\int \sqrt{3-2x-x^2}\,dx &= \int \sqrt{-((x+1)^2-4)}\,dx \\
&=\int \sqrt{4-(x+1)^2}\,dx  \\
\end{align}$$
Then I set $(x+1)=2\sin(t)$, $dx = 2\cos(t)\,dt$
$$\begin{align}
\int \sqrt{4-(2\sin(t))^2}2\cos(t)\,dt &= \int\sqrt{4-4\sin^2(t)}\,\,2\cos(t)\,dt \\
&= \int4\sqrt{1-\sin^2(t)}\cos(t)\,dt \\
&=4\int \sqrt{1-\sin^2(t)}\cos(t)\,dt  \\
&=4\int\cos^2(t)\,dt  \\
&=4\int\frac{1+\cos(2t)}\,2\,dt  \\
&=2 \int 1+\cos(2t) \,dt  \\
&=2t + \sin(2t) + C
\end{align}$$
$x+1 = 2\sin(t) \Leftrightarrow \frac{x+1}{2} = \sin(t) \Leftrightarrow t = \arcsin(\frac{x+1}{2})$
Replacing,
$$2\arcsin\left(\frac{x+1}{2}\right) + \sin\left(2\arcsin\left(\frac{x+1}{2}\right)\right) + C$$
But my book's solution is $$\frac{x+1}{2}\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)$$
So... what went wrong?

HINT:
Note that $\sin(2x)=2\sin(x)\cos(x)$, $\sin(\arcsin(x))=x$, and $\cos(\arcsin(x))=\sqrt{1-x^2}$.
Can you reconcile now?
