Notes: integration of 1/t^3 square root of t^2-1 From square root of 2 to 2

Friday, October 23, 2015

We'll start evaluating the integral, using
substitution.

Let sqrt(t^2 - 1) =
u.

We'll differentiate both
sides:

2tdt/2sqrt(t^2 - 1) =
du

du =tdt/sqrt(t^2 - 1)

We'll
determine t:

u = sqrt(t^2 -
1)

u^2 = t^2 - 1 => t^2 = u^2 +
1

t^4 = (u^2 + 1)^2

We'll
re-write the integral:

$\displaystyle\int$ dt/t^3*sqrt(t^2-1) = $\displaystyle\int$
tdt/t^4*sqrt(t^2 - 1)

$\displaystyle\int$ tdt/t^4*sqrt(t^2 - 1) = $\displaystyle\int$
du/(u^2 + 1)^2

$\displaystyle\int$ du/(u^2 + 1)^2 = $\displaystyle\int$ (u^2 + 1 -
u^2)dt/(u^2 + 1)^2

$\displaystyle\int$ (u^2 + 1 - u^2)dt/(u^2 + 1)^2 =
$\displaystyle\int$ dt - $\displaystyle\int$ u^2dt//(u^2 + 1)^2

We'll integrate the
last integral by parts, using the formula:

$\displaystyle\int$ f*g' = f*g
- $\displaystyle\int$ f'*g

Let f = u => f' =
du

Let g' = u/(u^2 + 1)^2 => g = $\displaystyle\int$ udu/(u^2 +
1)^2

We'll determine g using
substitution:

u^2 + 1 = v => 2udu = dv => udu
= dv/2

g = $\displaystyle\int$ dv/2v^2 = -1/2(u^2 +
1)

We'll apply the formula of integrating by
parts:

$\displaystyle\int$ u^2dt//(u^2 + 1)^2 = -u/2(u^2 + 1) + $\displaystyle\int$
du/2(u^2 + 1)

$\displaystyle\int$ u^2dt//(u^2 + 1)^2 = -u/2(u^2 + 1) +
(1/2)*arctan u + C

The result of integration of the
function is:

$\displaystyle\int$ dt/t^3*sqrt(t^2-1) = sqrt(t^2 - 1)
+ sqrt(t^2 - 1)/2t^2 - (1/2)*arctan [sqrt(t^2 - 1)] + C

To
determine the value of the definite integral, we'll use Leibniz Newton
formula:

$\displaystyle\int$
dt/t^3*sqrt(t^2-1) = F(2) -
F(sqrt2)

F(sqrt2) = sqrt(2 - 1) + sqrt(2 - 1)/4 -
(1/2)*arctan [sqrt(2 - 1)]

F(sqrt2) = 1 + 1/4 - ( $\displaystyle\pi$
/8)

F(2) = sqrt(2^2 - 1) + sqrt(2^2 - 1)/8 - (1/2)*arctan
[sqrt(2^2 - 1)]

F(2) = sqrt 3+ sqrt 3/8 - (1/2)*arctan
[sqrt3]

$\displaystyle\int$
dt/t^3*sqrt(t^2-1) = F(2) - F(sqrt2) =
sqrt 3+ sqrt 3/8 - (1/2)*arctan [sqrt3]- 5/4 + $\displaystyle\pi$ /
8

Notes