Question

How do you integrate `int sqrt(x^2-25)/x` using trig substitutions?

Answer 1

`int(sqrt(x^2-25))/xdx`

`color(red)(x=5secu)`

`color(red)((dx)/(du)=5secutanu`

`I=int(sqrt(25sec^2u-25))/(5secu) xx5secutanudu`

`I=int(sqrt(25(sec^2u-1)))/(5secu)xx5secutanudu`

now `color(red)(sec^2u=1+tan^2u)`

`:.color(red)(sec^2u-1=tan^2u)`

`=>I=int(sqrt(25tan^2u)/(5secu))xx5secutanudu`

`I=int(5tanu)/(5secu)xx5secutanudu`

now simplifying the algebra.

`I=int(cancel(5)tanu)/(cancel(5)cancel(secu))xx5cancel(secu)tanudu`

`I=5int(tan^2u)du`

using `color(red)(sec^2u=1+tan^2u)` once more.

`I=5int(sec^2u-1)du`

`I=5tanu-5u+C`

but

`color(red)(x=5secu=>u=sec^-1(x/5))`

`:. I=5tan(sec^-1(x/5))-5sec^-1x+C`

If you are used to hyperbolic functions the substitution

`x=5coshu` would make it less messy!