# What's my mistake in this integration??? $\int \sqrt{3-2x-x^2} dx$

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!

• Why do you think your answer is wrong? Sep 15, 2016 at 22:54

Answer is correct, if you want it to match with one that Wolfram produce, there should be done following: $$\sin(2\theta)=2\sin(\theta)\cos(\theta)=2\frac{1}{2}\sqrt{2^2-u^2}\frac{u}{2}=\frac{u}{2}\sqrt{4-u^2}\\ u\to x+1\\ \sin(2\theta)=\frac{1}{2}(x+1)\sqrt{3-2x-x^2}$$ And for second addendum $$\arccos(u/2)=\frac{\pi}{2}-\arcsin(u/2)\\ 2\arccos(u/2)=\pi-2\arcsin\left(\frac{x+1}{2}\right)$$ Altogether (hiding $\pi/2$ in constant): $$\int \sqrt{3-2x-x^2} dx = \frac{1}{2}(x+1)\sqrt{3-2x-x^2}+2\arcsin\left(\frac{x+1}{2}\right)+const$$