Question

How do you find the antiderivative of `int sqrt(x^2-1) dx`?

Answer 1

`= 1/2 x sqrt(x^2 - 1) - 1/2 cosh^(-1) (x) + C`

Explanation:

When you say you want the "anti-derivative" of `int sqrt(x^2-1) dx`, I am assuming you mean you want the "anti-derivative" that is `int sqrt(x^2-1) dx`.

Otherwise you are looking for: `int ( int sqrt(x^2-1) dx) \ dx`.

For `int sqrt(x^2-1) \ dx`, there is a convenient hyperbolic relationship:

`cosh^2 z - sin^2 z = 1`

If we let `x = cosh z, dx = sinh z \ dz`, we get this:

`int sqrt(cosh^2 z -1) sinh z \ dz`

`int sqrt(sinh^2 z) sinh z \ dz`

`= int sinh^2 z \ dz`

By the hyperbolic double angle formula (`cosh 2z = 1 + 2 sinh^2 z`):

`=1/2 int cosh 2z - 1\ dz`

`=1/2 ( 1/2 sinh 2z - z) + C = 1/4 sinh 2z - 1/2 z + C qquad triangle`

Now:

`sinh 2z = 2 sinh z cosh z =`

`= 2 sqrt(x^2 - 1) * x`

So `triangle` becomes:

`= 1/2 x sqrt(x^2 - 1) - 1/2 cosh^(-1) (x) + C`

NB

If you like we can take `cosh^(-1) (x)` and move it from hyperbolic to natural log status:

let `y = cosh^(-1) (x)`

`x = cosh y`

`= (e^y + e^(-y))/2`

`implies x = (e^y + e^(-y))/2`

`implies 2 x e^ y = e^(2y) + 1`

`implies (e^(y))^2 - 2 x e^ y + 1 = 0`

From the quadratic formula:

`e^y = (2x pm sqrt(4x^2 - 4 ))/(2) = x pm sqrt(x^2 - 1 )`

`implies y = ln ( x pm sqrt(x^2 - 1 ))`

So `triangle` becomes:

`= 1/2 x sqrt(x^2 - 1) - 1/2 ln ( x pm sqrt(x^2 - 1 )) + C`

Answer 2

`intsqrt(x^2-1)` `dx=1/2(xsqrt(x^2-1)-ln|x+sqrt(x^2-1)|)+"c"`

Explanation:

We are trying to evaluate `intsqrt(x^2-1)` `dx`.

Firstly, we should let `x=secu`

Then our integrand becomes `sqrt(sec^2u-1)=tanu` and

`dx/(du)=secutanu rArrdx=secutanu` `du`

`intsqrt(x^2-1)` `dx=intsecutan^2u` `du=`
`intsecu(sec^2u-1)` `du=intsec^3u-secu` `du=`
`intsec^3u` `du-intsecu` `du`

The integral of `secu` is a standard and is evaluated as `ln|secu+tanu|+"C"`.

The integral of `sec^3u` will require integration by parts. If you wish to know how to do it, I will add it later. If you already do, just follow the working.

`intsec^3u=1/2secutanu+1/2ln|secu+tanu|+"C"`

`intsec^3u` `du-intsecu` `du=`
`1/2secutanu+1/2ln|secu+tanu|+"C"-ln|secu+tanu|+"C"`
`=1/2secutanu-1/2ln|secu+tanu|+"c"`

Now all that's left is to rewrite the answer in terms of `x`.

`1/2secutanu-1/2ln|secu+tanu|+"c"=`

`1/2xsqrt(x^2-1)-1/2ln|x+sqrt(x^2-1)|+"c"`

How to evaluate `intsec^3u` `du`:

`intsec^3u` `du=intsec^2usecu` `du`

`f(x)=secu rArr f'(x)=secutanu`
`g'(x)=sec^2u rArr g(x)=tanu`

`intsec^2usecu` `du=secutanu-intsecutan^2u` `du=`
`secutanu-intsecu(sec^2u-1)` `du=`
`secutanu-(intsec^3u` `du-intsecu` `du)=`

`intsec^3u` `du=secutanu-intsec^3u` `du+intsecu` `du`

`2intsec^3u` `du=secutanu+ln|secu+tanu|`

`intsec^3u` `du=1/2(secutanu+ln|secu+tanu|)+"C"`