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

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Answer 1 · 2016-02-17T10:54:32

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

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:

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

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