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

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Answer 1 · 2018-07-04T14:41:29

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

#\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 · 2018-07-04T14:41:50

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

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#