How do you integrate #int (x^2-8x+21)^(3/2)# using trig substitutions?
1 Answer
Use the substitution
Explanation:
Let
#I=int(x^2−8x+21)^(3/2)dx#
Complete the square:
#I=int((x-4)^2+5)^(3/2)dx#
Apply the substitution
#I=25intsec^5thetad theta#
Apply the integration by parts:
#u(theta)=sec^3theta# ,#u'(theta)=3sec^3thetatantheta#
#v'(theta)=sec^2theta# ,#v(theta)=tantheta#
Hence:
#intsec^5thetad theta=sec^3thetatantheta-3intsec^3thetatan^2thetad theta#
Rearrange:
#intsec^5thetad theta=1/4sec^3thetatantheta+3/4intsec^3thetad theta#
Similarly:
#intsec^3thetad theta=1/2secthetatantheta+1/2ln|sectheta+tantheta|+C#
Hence
#I=25{1/4sec^3thetatantheta+3/8secthetatantheta+3/8ln|sectheta+tantheta|}+C#
Reverse the substitution:
#I=1/8(x-4)(2x^2-16x+57)sqrt(x^2-8x+21)+75/8ln|x-4+sqrt(x^2-8x+21)|+C#