How do you integrate int (x^2-8x+21)^(3/2) using trig substitutions?
1 Answer
Jun 3, 2018
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