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

Question

Charles Arocha · Accepted Answer

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

Charles Anctil · Answer

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#

Aurora Alvarado · Answer

undefined To find the integral of $ \frac{1}{\sqrt{-x^2 - 4x}} $, we can start by completing the square in the denominator to simplify the expression. Then, we can use a trigonometric substitution. Let $ u = -x - 2 $. This substitution will help simplify the integral. After performing the substitution, you'll end up with an integral involving trigonometric functions. Finally, you can use trigonometric identities and integration techniques to solve the integral.