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

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### 1 Answer

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:

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

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

`int` du/(u^2 + 1)^2 = `int` (u^2 + 1 - u^2)dt/(u^2 + 1)^2

`int` (u^2 + 1 - u^2)dt/(u^2 + 1)^2 = `int` dt - `int` u^2dt//(u^2 + 1)^2

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

`int` f*g' = f*g - `int` f'*g

Let f = u => f' = du

Let g' = u/(u^2 + 1)^2 => g = `int` udu/(u^2 + 1)^2

We'll determine g using substitution:

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

g = `int` dv/2v^2 = -1/2(u^2 + 1)

We'll apply the formula of integrating by parts:

`int` u^2dt//(u^2 + 1)^2 = -u/2(u^2 + 1) + `int` du/2(u^2 + 1)

`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:

`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:

**`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 - (`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]

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