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How do you integrate #int 1/sqrt(9x^2-18x+8) # using trigonometric substitution?

1 Answer

Mar 9, 2016

#1/3 cosh^-1 (3x-3) + C#

Explanation:

Since: #9x^2 - 18x + 8 = (3x-3)^2 - 1#

We use the following substitution:

#3x -3 = cosh u#
#dx = 1/3 (sinh u ) du#

Hence:
#sqrt(9x^2 - 18x + 8) = sqrt((3x-3)^2 - 1) = sqrt(cosh^2 u - 1) = sqrt(sinh^2 u) = sinh u#

So we have:

#int 1/sqrt(9x^2 - 18x + 8) dx = int 1/sinhu * 1/3 sinhu * du = 1/3u + C = 1/3 cosh^-1 (3x-3) + C#

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