Question

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

Answer 1

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`