Question

Can you help me integrate this, its something to do with hyperbolics?

`int_0.5^1.3sqrt(4x^2 -1)dx`

Answer 1

The answer is `=1.16`

Explanation:

First compute the indefinite integral

Let `2x=cosh(u)`, `=>`, `2dx=sinh(u)du`

`sinh(u)=sqrt(cosh^2(u)-1)`

Therefore,

`intsqrt(4x^2-1)dx=1/2intsqrt(cosh^2u-1)*sinhudu`

`=1/2intsinh^2udu`

But,

`cosh2u=1+2sinh^2u`

`sinh^2u=1/2(cosh2u-1)`

Therefore,

`intsqrt(4x^2-1)dx=1/2int(1/2(cosh2u-1))du`

`=1/8sinh2u-1/4u`

`=1/8*2*sqrt(4x^2-1)*2x-1/4arccosh(2x)+C`

`=1/2xsqrt(4x^2-1)-1/4arccosh(2x)+C`

Now, compute the definite integral

`int_0.5^1.3sqrt(4x^2-1)dx=[1/2xsqrt(4x^2-1)-1/4arccosh(2x)]_0.5^1.3`

`=(1/2*1.3sqrt(4*1.3^2-1)-1/4arccosh(2*1.3))-(1/2*0.5sqrt(4*0.5^2-1)-1/4arccosh(2*0.5))`

`=1.16`

Answer 2

`int_0.5^1.3 sqrt(4x^2-1) dx = 1.56-1/4 cosh^(-1) 2.6 ~~ 1.15764`

Explanation:

Note that:

`sinh^2 t = cosh^2 t - 1 = 1/2 (-1 + cosh 2t)`

`sinh 2t = 2sinh t cosh t`

So let's try a hyperbolic substitution:

`x = 1/2 cosh u`

`dx/(du) = 1/2 sinh u`

Note that for the given integration range we have `sinh u >= 0`, so we can assume `sqrt(cosh^2 u - 1) = sinh u`

The `x` value `0.5` corresponds to `cosh u = 1` and hence `u=0`.

The `x` value `1.3` corresponds to `cosh u = 2.6` and hence `u = cosh^(-1) (2.6)`

Note that:

`sinh(cosh^(-1) 2.6) = sqrt((2.6)^2 - 1) = sqrt(6.76-1) = sqrt(5.76) = 2.4`

Then:

`int_0.5^1.3 sqrt(4x^2-1) dx = int_0^(cosh^(-1) 2.6) sqrt(cosh^2 u-1) dx/(du) du`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = int_0^(cosh^(-1) 2.6) 1/2 sinh^2 u color(white)(.)du`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = int_0^(cosh^(-1) 2.6) 1/4 (-1+cosh 2u) color(white)(.)du`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = [-1/4 u + 1/8 sinh 2u]_0^(cosh^(-1) 2.6)`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = -1/4 cosh^(-1) 2.6 + 1/8 sinh 2(cosh^(-1)(2.6))`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = -1/4 cosh^(-1) 2.6 + 1/4 sinh (cosh^(-1)(2.6)) cosh (cosh^(-1)(2.6))`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = -1/4 cosh^(-1) 2.6 + 1/4 (2.4)(2.6)`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) = 1.56-1/4 cosh^(-1) 2.6`

`color(white)(int_0.5^1.3 sqrt(4x^2-1) dx) ~~ 1.15764`