`int_4^5(sqrt(x^2-16))/x^2 dx`

This question contains a square root which is in the form of the 3rd substitution suggestion given at the top, that is:

For `sqrt(x^2-a^2)`, use `x=a\ sec theta`

So we have `a=4` and we let

`x = 4 sec θ`

and this gives

`x^2= 16 sec^2 θ`

and

`dx = 4 sec θ tan θ\ d θ`

Simplifying the square root part:

`sqrt(x^2-16)=sqrt(16 sec^2 theta-16)`

`=sqrt(16(sec^2theta-1))`

`=sqrt16sqrt(tan^2theta)`

`=4 tan theta`

Substituting `dx = 4 sec theta tan theta\ d theta,` `x^2= 16 sec^2 theta` and `sqrt(x^2-16)=4 tan theta` into the given integral gives us the following. (We take the indefinite case first and then do the substitution of upper and lower limits later, to make the writing a bit easier.)

`int(sqrt(x^2-16))/x^2dx=int((4 tan theta))/(16 sec^2 theta)(4 sec theta tan theta) d theta`

`=int(16 tan^2 theta sec theta)/(16 sec^2 theta) d theta`

`=int(tan^2 theta)/(sec theta) d theta`

`=int(sec^2theta-1)/(sec theta) d theta`

`=int((sec^2 theta)/(sec theta)-1/(sec theta)) d theta`

`=int (sec theta-cos theta) d theta`

`=[ln |sec theta+tan theta|-sin theta]+K`

Now, our question was a definite integral, so we need to either re-express our answer in terms of the original variable , `x`, or we could work it using `theta`.

#### Changing back to `x`

Earlier, we let `x = 4 sec θ`, so we get `sec theta=x/4`.

Using a triangle like we did earlier, we can also derive that:

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

and

`tan theta=(sqrt(x^2-16))/4`

Therefore, we can conclude that:

`int_4^5(sqrt(x^2-16))/x^2 dx=[ln|sec theta+tan theta|-sin theta]_(theta=?)^(theta=?)`

`=[ln|x/4+(sqrt(x^2-16))/4|-(sqrt(x^2-16))/x]_4^5`

`=[ln|5/4+3/4|-3/5]-[ln|1+0|-0]`

`=[ln|2|-3/5]-[0]`

`=0.09315`

#### Leaving it in terms of `theta`

Since `sec theta=x/4`, then as `x` ranges from `4` to `5`, then `sec theta` will range from `1` to `1.25`.

So the required upper and lower limits for `theta` (these are the missing question mark "`theta=?`" values in the above answer) will be

`theta="arcsec"(1)= 0`

and

`theta="arcsec"(1.25)=0.6435011`

Returning to our answer in `theta`, and substituting our upper and lower values gives:

`[ln |sec theta+tan theta|-sin theta]_(theta=0)^(theta=0.6435011)`

`=[ln |sec 0.6435011+tan 0.6435011|-sin 0.6435011]-[ln |sec 0+tan 0|-sin 0]`

`=0.09315`, which is the same as our earlier answer.

In this example, both approaches (leaving it in terms of `theta` or changing back to `x`) is about the same amount of work.