Integrate using trigo substitution #int dx/(sqrt(x^2-4x))^3# ?

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Answer 1 · 2018-07-14T18:45:17

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

At first we Substitute #u=x-2,du=dx#

and we get

#int1/(u^2-4)^(3/2)du#
now we substitue #u=2sec(s),du=2tan(s)sec(s)ds#

then we get
#1/4int cot(s)csc(s)ds=-csc(s)/4+C#
with

#s=sec(-1)(u/2)# we get

#-1/4csc(sec^(-1)(u/2))+C#
note that

#csc(sec^(-1)(z))=1/sqrt(1-1/z^2)# then we get

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

we get the result
#(2-x)/(4sqrt((x-4)x)+C#