Question

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

Answer 1

`\int \frac{1}{\sqrt{-x^2-4x}}dx=\sin^{-1}(\frac{x+2}{2})+C`

Explanation:

`\int \frac{1}{\sqrt{-x^2-4x}}dx`

`=\int \frac{1}{\sqrt{-x^2-4x-4+4}}dx`

`=\int \frac{1}{\sqrt{4-(x^2+4x+4)}}dx`

`=\int \frac{1}{\sqrt{4-(x+2)^2}}dx`

Let `x+2=2\sin\theta\implies dx=2\cos\theta\ d\theta`

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

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

`=\int \frac{\cos\theta\d\theta}{\cos\theta}`

`=\int \ d\theta`

`=\theta+C`

`=\sin^{-1}({x+2}/2)+C`

Answer 2

The answer is `=arcsin((x+2)/2)+C`

Explanation:

The denominator is

`sqrt(-x^2-4x)=sqrt(4-(x+2)^2)`

Therefore, the integral is

`I=int(dx)/(sqrt(-x^2-4x))=int(dx)/sqrt(4-(x+2)^2)`

Let `u=(x+2)/2`

`=>`, `du=1/2dx`

Therefore,

`I=int(2du)/(sqrt(4-4u^2))= int(du)/(sqrt(1-u^2))`

Let `u=sintheta`, `=>`, `du=costhetad theta`

The integral is

`I=int(costhetad theta)/costheta`

`=intd theta`

`=theta`

`=arcsinu`

`=arcsin((x+2)/2)+C`