`int_sqrt(2)^2 1/(t^2 sqrt(t^2 - 1)) dt` Evaluate the integral

You need to re-write the expression` t^2 - 1` , such that:

`t^2 - 1 = t^2(1 - (1/t)^2)`

You need to use the following substitution, such that:

`1/t = sin u => -1/(t^2) dt = cos u du => (dt)/(t^2) = -cos u du`

`u = arcsin (1/t)`

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = - int_(u_1)^(u_2) (cos u du)/(1/(sin u)*sqrt(1 - sin^2 u))`

You need to use the basic trigonometric formula `1 - sin^2 u = cos^2 u,` such that:

`- int_(u_1)^(u_2) (cos u*sin u du)/(sqrt(1 - sin^2 u)) = - int_(u_1)^(u_2) `

`- int_(u_1)^(u_2) (cos u*sin u du)/(sqrt(cos^2 u)) = - int_(u_1)^(u_2) (cos u*sin u du)/(cos u)`

Reducing like terms yields:

`-int_(u_1)^(u_2) (sin udu) = -(-cos u)|_(u_1)^(u_2)`

Replacing back the variable, yields:

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) =(cos (arcsin (1/t)))_(sqrt 2)^2`

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) =cos (arcsin (1/2)) - cos (arcsin (1/sqrt 2))`

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = cos(pi/6) - cos(pi/4)`

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = sqrt3/2 - sqrt2/2`

`int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = (sqrt3 - sqrt2)/2`

Hence, evaluating the given integral yields `int_(sqrt 2)^2 (dt)/(t^3*sqrt(t^2-1)) = (sqrt3 - sqrt2)/2.`