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

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Answer 1 · 2016-10-21T11:41:47

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

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

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

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

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

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!

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