How do you integrate #int 1/sqrt(4x^2-12x+8) # using trigonometric substitution?

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Answer 1 · 2018-03-25T19:31:43

#int1/sqrt(4x^2-12x+8)dx=1/2"arcosh"(2x-3)+"c"#

We want to find #int1/sqrt(4x^2-12x+8)dx#

We start by transforming the integral

#int1/sqrt(4x^2-12x+8)dx=int1/sqrt(4x^2-12x+9-1)dx=int1/sqrt((2x-3)^2-1)dx#.

Now let #u=2x-3# and #du=2dx#. The integral becomes

#int1/sqrt((2x-3)^2-1)dx=1/2int1/sqrt(u^2-1)du#

The integral is a standard integral and evaluates to

#1/2int1/sqrt(u^2-1)du=1/2"arcosh"u + "c"=1/2"arcosh"(2x-3)+"c"#