This is how I'm trying to integrate this function:
\begin{align*} \int \sqrt{3-2x-x^2}\, dx &= \int \sqrt{4-(x+1)^2}\, dx \\ &= \int \sqrt{2^2-(x+1)^2}\, dx \end{align*} Here I make the substitution:
$$ u=x+1 $$ $$du=dx$$
So the integral is now:
$$ \int \sqrt{2^2-u^2}\, du $$
I make a trigonometric substitution thinking about a right triangle where the hypothenuse is $2$, the adjacent side is $u$, the opposite side is $\sqrt{2^2-u^2}$, and the angle is called $\theta$.
$$ \sin(\theta)= \frac{\sqrt{2^2-u^2}}{ 2}$$ $$\bbox[2px,border:2px solid red] { 2\sin(\theta)= \sqrt{2^2-u^2}\qquad }$$
$$\frac{u}{2} =\cos(\theta)$$ $$ u=2 \cos(\theta)$$ $$ \bbox[2px,border:2px solid red] {du=-2 \sin(\theta)\,d\theta\qquad }$$
So I write the integral as: \begin{align*} \int 2\sin(\theta)(-2)\sin(\theta)\,d\theta &= \int (-4)\sin(\theta)\sin(\theta)\,d\theta \\ &= \int (-4){\sin}^2(\theta)\,d\theta \\ &= (-4)\int {\sin}^2(\theta)\,d\theta \\ &= (-4)\int \frac{1}{2}(1-\cos(2\theta))\,d\theta \\ &= (-4)\frac{1}{2}\int (1-\cos(2\theta))\,d\theta \\ &= (-2)\int (1-cos(2\theta))\,d\theta \\ &= (-2)\left[\int d\theta-\int \cos(2\theta)\,d\theta \right]\\ &= (-2)\left[\theta-\int \cos(2\theta)\,d\theta \right]\\ &= (-2)\left[\theta-\frac{1}{2} \sin(2\theta) \right]\\ &= \sin(2\theta) -2\theta \end{align*} And since $\cos(\theta) = u/2$, I know that $\theta =\arccos(u/2)$.
Therefore I have:
$$ \sin(2\arccos(u/2)) -2\arccos(u/2)$$
According to my first substitution $u=x+1$ so the final result is:
$$\bbox[2px,border:2px solid red] {\sin\left(2\arccos\left(\frac{x+1}{2}\right)\right) -2\arccos\left(\frac{x+1}{2}\right) + constant }\qquad$$
Can anyone help me? I don't understand what I'm doing wrong. Thanks!
