`int1/(t^2sqrt(t^2-16))dt`

Let's evaluate the integral by applying integral substitution,

Let `t=4sec(u)`

`=>dt=4tan(u)sec(u)du`

Plug the above in the integral,

`int1/(t^2sqrt(t^2-16))dt=int(4tan(u)sec(u))/((4^2sec^2(u)sqrt(4^2sec^2(u)-16)))du`

`=int(4tan(u)sec(u))/(16sec^2(u)*4sqrt(sec^2(u)-1))du`

`=inttan(u)/(16sec(u)sqrt(sec^2(u)-1))du`

Now use the identity:`sec^2(x)=1+tan^2(x)`

`=1/16inttan(u)/(sec(u)sqrt(1+tan^2(u)-1))du`

`=1/16inttan(u)/(sec(u)tan(u))du`

`=1/16int1/sec(u)du`

`=1/16intcos(u)du`

Now use the standard integral:`intcos(x)dx=sin(x)+C`

`=1/16sin(u)+C`

We have used `t=4sec(u)`

`=>u=arcsec(t/4)`

Substitute back u in the solution,

`=1/16sin(arcsec(t/4))+C`

Simplify the above by assuming the right triangle with angle `theta=arcsec(t/4)`

Hypotenuse is t and adjacent side is 4, Opposite side(O) can be found by using pythagorean identity,

`4^2+O^2=t^2`

`O^2=t^2-16`

`O=sqrt(t^2-16)`

`sin(theta)=sqrt(t^2-16)/t`

Hence the solution is `1/16(sqrt(t^2-16)/t)+C`

You need to re-write the expression `t^2 - 16` , such that:

`t^2 - 16 = t^2(1 - (4/t)^2)`

You need to use the following substitution, such that:

`4/t = sin u => -4/(t^2) dt = cos u du => (dt)/(t^2) = -(1/4)*cos u du`

`u = arcsin (4/t)`

`int (dt)/(t^2*sqrt(t^2-16)) = (-1/4) int (cos u du)/(4/(sin u)*sqrt(1 - sin^2 u))`

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

`(-1/16) int (cos u*sin u du)/(sqrt(1 - sin^2 u)) = (-1/16) int (cos u*sin u du)/(sqrt(cos^2 u))`

`(-1/16) int (cos u*sin udu)/(sqrt(cos^2 u)) = (-1/16) int (cos u*sin udu)/(cos u)`

Reducing like terms yields:

`(-1/16) int (sin udu) = (-1/16)(-cos u) + c`

Replacing back the variable, yields:

`int (dt)/(t^2*sqrt(t^2-16)) = (1/16)(cos (arcsin (4/t))) + c`

Hence, evaluating the given integral yields `int (dt)/(t^2*sqrt(t^2-16)) = (1/16)(cos (arcsin (4/t))) + c.`