`int (dt)/(t^2 sqrt(t^2 - 16))` Evaluate the integral

`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`