How do you integrate #int 1/sqrt(9x^2-18x+18) # using trigonometric substitution?

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Answer 1 · 2016-03-01T13:57:23

#=1/3sinh^-1(x-1)+C#

First of all, we complete the square/ re write in vertex form the quadratic under the square root:

#9x^2-18x+18#
#=9(x^2-2)+18#
#=9(x-1)^2+9#

So the integral can be re written as:

#int1/sqrt(9(x-1)^2+9)dx=1/3int1/sqrt((x-1)^2+1)dx#

(Note the 9 under the square root has been factored out to the front)

Now consider the substitution #sinh(u)=x-1#
It will follow that #cosh(u)du=dx#

Now substitute this into the integral to get:

#1/3intcosh(u)/sqrt(sinh^2(u)+1)dx#

Use the identity: #cosh^2(u)-sinh^2(u)=1# to re write the bottom as:

#1/3intcosh(u)/sqrt(cosh^2(u))dx=1/3intcosh(u)/cosh(u)du = 1/3intdu#

Now evaluating the integral and reversing the substitution:

#1/3u +C#
#=1/3sinh^-1(x-1)+C#