How do you integrate #int e^(2x)/sqrt(e^(2x) -36)dx# using trigonometric substitution?

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Answer 1 · 2016-03-10T22:41:16

#sqrt(e^(2x)-36) + C#

We don't really need substitution since we have a perfect form
#int (f'(x))/(2sqrt(f(x))) ->sqrt(f(x))#
but here you go:

Hence:

#e^(2x) - 36 = (6cosh u)^2 - 36 = 36(cosh^2 u - 1)#
#36sinh^2 u= (6sinh u)^2#

So we have:

#int e^(2x)/sqrt(e^(2x) -36)dx = int e^x/sqrt(e^(2x) -36) (e^xdx) = int (6coshu)/(6 sinh u) (6sinhu du) = int 6coshu du = 6sinh u + C#

To substitute back we use:
#sinhu = sqrt(cosh^2u -1 ) = sqrt(e^(2x)/36-1) = sqrt(e^(2x)-36)/6#

Finally:

#int e^(2x)/sqrt(e^(2x) -36)dx=sqrt(e^(2x)-36) + C#