Question

How do you integrate `int x/sqrt(x^2-25)` by trigonometric substitution?

Answer 1

`int x/sqrt(x^2-25)dx = sqrt(x^2-25) + C`

Explanation:

You do not need a trigonometric substitution to solve this integral.

Substituting:

`t=x^2-25`
`dt= 2xdx`

you have:

`int x/sqrt(x^2-25)dx = 1/2 int dt/sqrt(t) = sqrt(t)+C`

and substituting back `x`:

`int x/sqrt(x^2-25)dx = sqrt(x^2-25) + C`

Answer 2

As has been shown, trigonometric substitution is a waste of time and effort, but it can be used with the substitution `x=5sectheta=>dx=5secthetatanthetad theta`.

`intx/sqrt(x^2-25)dx=int(5sectheta)/sqrt(25sec^2theta-25)(5secthetatanthetad theta)`

Note that `sqrt(25sec^2theta-25)=5sqrt(sec^2theta-1)=5tantheta` via the form of the Pythagorean identity `tan^2theta+1=sec^2theta`.

`=int(5sectheta)/(5tantheta)(5secthetatanthetad theta)=5intsec^2thetad theta=5tantheta+C`

Note that `tantheta=sqrt(sec^2theta-1)` and from our original substitution `sectheta=x/5`:

`=5sqrt(sec^2theta-1)+C=5sqrt(x^2/25-1)+C=5sqrt((x^2-25)/25)+C`

`=sqrt(x^2+25)+C`