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

1 Answer
Jun 3, 2018

Use the substitution x-4=sqrt5tantheta then apply integration by parts.

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 x-4=sqrt5tantheta:

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