How do you evaluate the integral #int x^2arcsecx#?

1 Answer
Jun 22, 2017

#1/3x^3sec^-1(x)-1/6xsqrt(x^2-1)-1/6lnabs(x+sqrt(x^2-1))+C#

Explanation:

#I=intx^2sec^-1(x)dx#

Let #x=sectheta#. Then, #theta=sec^-1(x)# and #dx=secthetatanthetad theta#.

#I=intsec^2theta(theta)(secthetatanthetad theta)=intthetasec^3thetatanthetad theta#

Note that #intsec^3thetatanthetad theta=intsec^2thetacolor(white).d(sec theta)=1/3sec^3theta#. Then, we can easily perform integration by parts on #I# by letting:

#u=theta" "=>" "du=d theta#

#dv=sec^3thetatanthetad theta" "=>" "v=1/3sec^3theta#

So:

#I=1/3thetasec^3theta-1/3intsec^3thetad theta" "" "" "color(red)star#

We can solve this integral with integration by parts again, now letting:

#u=sectheta" "=>" "du=secthetatantheta d theta#

#dv=sec^2thetad theta" "=>" "v=tantheta#

So:

#I=1/3thetasec^3theta-1/3(secthetatantheta-intsecthetatan^2thetad theta)#

Letting #tan^2theta=sec^2theta-1# this becomes:

#I=1/3thetasec^3theta-1/3secthetatantheta+1/3intsectheta(sec^2theta-1)#

#color(white)I=1/3thetasec^3theta-1/3secthetatantheta+1/3intsec^3thetad theta-1/3intsecthetad theta#

#color(white)I=1/3thetasec^3theta-1/3secthetatantheta-1/3lnabs(sectheta+tantheta)+1/3intsec^3thetad theta#

Earlier (#color(red)star#) we noted that #I=1/3thetasec^3theta-1/3intsec^3thetad theta#, so we can set this to be equivalent to the expression we just found.

#1/3thetasec^3theta-1/3intsec^3thetad theta=1/3thetasec^3theta-1/3secthetatantheta-1/3lnabs(sectheta+tantheta)+1/3intsec^3thetad theta#

Rearranging:

#-2/3intsec^3thetad theta=-1/3secthetatantheta-1/3lnabs(sectheta+tantheta)#

Which gives:

#-1/3intsec^3thetad theta=-1/6secthetatantheta-1/6lnabs(sectheta+tantheta)#

Plugging this into #color(red)star# yields:

#I=1/3thetasec^3theta-1/6secthetatantheta-1/6lnabs(sectheta+tantheta)#

We can now return to our original variable #x# using #x=sectheta#, #tantheta=sqrt(sec^2theta-1)=sqrt(x^2-1)# and #theta=sec^-1(x)#:

#I=1/3x^3sec^-1(x)-1/6xsqrt(x^2-1)-1/6lnabs(x+sqrt(x^2-1))+C#