How do you integrate #int 1/(x^2sqrt(4+x^2))# by trigonometric substitution?

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Answer 1 · 2018-04-08T12:01:39

The answer is #=-sqrt(4+x^2)/(4x)+C#

Perform this integral by substitution

Let #x=2tanu#, #=>#, #dx=2sec^2udu#

#sqrt(4+x^2)=sqrt(4+4tan^2u)=2secu#

The integral is

#I=int(2sec^2udu)/(4tan^2u*2secu)#

#=1/4int(sec udu)/(tan^2u)#

#=1/4int(cosu du)/(sin^2u)#

Let #v=sinu#, #=>#, #dv=cosu du#

Therefore,

#I=1/4int(dv)/(v^2)#

#=1/4*-(1/v)#

#=-1/(4sinu)#

#=-1/(4*x/(sqrt(x^2+4)))#

#=-sqrt(4+x^2)/(4x)+C#