How do you integrate from -6 to-3 for #sqrt((x^2-9))/2#?

1 Answer
Apr 12, 2015

Method is this

Substitute #x= 3 sec theta#
For any variable substitution in the integration, Integrant, Limits and the integration operator need to be changed.

So Integrant is #sqrt(9sec^2theta-9)/2 = (3tantheta)/2#

The limits are #@x=-6 to x=-3# changes as well.

Upper Limit calculation
#3sectheta = -3#
#sec theta = -1#
# theta = pi#

Lower Limit calculation
#3sec theta = -6#
#sec theta = -2#
#theta = (5pi)/6#

Integral operator change
#dx = sectheta tantheta d theta#
#int_-6^-3 sqrt(x^2-9)/2 = int_((5pi)/6)^pi (3secthetatan^2theta d theta)/2#
# = 3/2int_((5pi)/6)^pi (sec^3theta - sec theta) d theta#

Check how to evaluate #int sec^3 theta d theta# here

#=3/2{1/2(sec theta tantheta+ln|sec theta+tantheta|)-ln|sec theta+tantheta|}_((5pi)/6)^pi#

#=3/4{sec theta tantheta-ln|sec theta+tantheta|}_((5pi)/6)^pi#
#=-0.0082#Calculation by me using Excel