# How do you find int 1/sqrt(-x^2-4x)?

$\setminus \int \setminus \frac{1}{\setminus \sqrt{- {x}^{2} - 4 x}} \mathrm{dx} = \setminus {\sin}^{- 1} \left(\setminus \frac{x + 2}{2}\right) + C$

#### Explanation:

$\setminus \int \setminus \frac{1}{\setminus \sqrt{- {x}^{2} - 4 x}} \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{- {x}^{2} - 4 x - 4 + 4}} \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{4 - \left({x}^{2} + 4 x + 4\right)}} \mathrm{dx}$

$= \setminus \int \setminus \frac{1}{\setminus \sqrt{4 - {\left(x + 2\right)}^{2}}} \mathrm{dx}$

Let $x + 2 = 2 \setminus \sin \setminus \theta \setminus \implies \mathrm{dx} = 2 \setminus \cos \setminus \theta \setminus d \setminus \theta$

$= \setminus \int \setminus \frac{2 \setminus \cos \setminus \theta \setminus d \setminus \theta}{\setminus \sqrt{4 - 4 \setminus {\sin}^{2} \setminus \theta}}$

$= \setminus \int \setminus \frac{2 \setminus \cos \setminus \theta \setminus d \setminus \theta}{2 \setminus \sqrt{1 - \setminus {\sin}^{2} \setminus \theta}}$

$= \setminus \int \setminus \frac{\setminus \cos \setminus \theta \setminus d \setminus \theta}{\setminus \cos \setminus \theta}$

$= \setminus \int \setminus d \setminus \theta$

$= \setminus \theta + C$

$= \setminus {\sin}^{- 1} \left(\frac{x + 2}{2}\right) + C$

Jul 4, 2018

The answer is $= \arcsin \left(\frac{x + 2}{2}\right) + C$

#### Explanation:

The denominator is

$\sqrt{- {x}^{2} - 4 x} = \sqrt{4 - {\left(x + 2\right)}^{2}}$

Therefore, the integral is

$I = \int \frac{\mathrm{dx}}{\sqrt{- {x}^{2} - 4 x}} = \int \frac{\mathrm{dx}}{\sqrt{4 - {\left(x + 2\right)}^{2}}}$

Let $u = \frac{x + 2}{2}$

$\implies$, $\mathrm{du} = \frac{1}{2} \mathrm{dx}$

Therefore,

$I = \int \frac{2 \mathrm{du}}{\sqrt{4 - 4 {u}^{2}}} = \int \frac{\mathrm{du}}{\sqrt{1 - {u}^{2}}}$

Let $u = \sin \theta$, $\implies$, $\mathrm{du} = \cos \theta d \theta$

The integral is

$I = \int \frac{\cos \theta d \theta}{\cos} \theta$

$= \int d \theta$

$= \theta$

$= \arcsin u$

$= \arcsin \left(\frac{x + 2}{2}\right) + C$