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

2 Answers
Jan 14, 2017

#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#

Jan 14, 2017

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#