How do you integrate #int 1/sqrt(9x^2-18x+13) # using trigonometric substitution?

1 Answer
Feb 17, 2016

#sinh^-1(3/2(x-1))+C#

Explanation:

I can give you a solution but it uses a hyperbolic substitution as oppose to a standard trig substitution but it is still very similar.

We begin by re writing the quadratic under the square root in completed square / vertex form.

#9x^2-18x+13=9(x^2-2x)+13#
#=9(x-1)^2+4#

We can re write the integral as:

#intdx/sqrt(9x^2-18x+13) = intdx/sqrt(9(x-1)^2+4)#

At this point we may like to consider the substitution:

#3(x-1) = 2sinh(u) -> 9(x-1)^2=4sinh^2(u)#

It also follows from this substitution that:

#3dx = 2cosh(u)du#

Notice it is not a standard trig function but a hyperbolic function

Now putting this into the integral we get:

#2/3intcosh(u)/sqrt(4sinh^2(u)+4)du#

We can factor #4# out of the square root on the bottom (leaving us with #2#) which we will put on the front of the integral:

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

From the hyperbolic identity: #cosh^2(u)-sinh^2(u) =1#
we can replace the expression under the square root with:

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

Which simplifies to:

#1/3intcosh(u)/cosh(u)du=1/3intdu=1/3u+C#

Now reverse the substitution for #u# and we get:

#1/3sinh^-1(3/2(x-1))+C#

You can leave it here or if you want to go a bit further, re-write this in terms of the definition of the #sinh# inverse function which would give:

#1/3ln{3/2(x-1)+sqrt(9/4(x-1)^2+1) }+C#