How do you integrate #int dx/sqrt(x^2+4)# using trig substitutions?

1 Answer
Roy E.
Dec 14, 2016

Substitute #x=2 sinh u# quickly getting the integral to be #sinh^{-1}(x/2)+C#

Explanation:

#x=2sinh u#, #dx=2cosh u du# giving the integral as
#int2coshu/ sqrt(4 sinh^2u+4)du#
#=int cosh u/ cosh u du# because #cosh^2u -sinh^2 u =1#
#=int1du#
#=u+C#
#=sinh^-1(x/2)+C#

Some people prefer to write this as #ln(x+sqrt(x^2+4))# which is equivalent, as it differs from the first solution by a fixed constant #ln(1/2)#.

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