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

Question

2749 views around the world

You can reuse this answer
Creative Commons License

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!