Question

How do you integrate `int sqrt((x+3)^2-100)` using trig substitutions?

Answer 1

The answer is `=(x+3)/2(sqrt(((x+3)^2-100)))-50ln((sqrt(((x+3)^2)-100)+(x+3))/10)+C`

Explanation:

Let `u=x+3` then `du=dx`
`int(sqrt((x+3)^2-100))dx=int(sqrt(u^2-100))du`
Then let `u=10sectheta` `=>` `du=10secthetatantheta`
`int(sqrt(u^2-100))du=int(sqrt(100sec^2theta-100))10secthetatantheta(d(theta))`
`=100intsecthetatan^2thetad(theta)`
`=100intsectheta(sec^2theta-1)d(theta)`
`=100int(sec^3theta-sectheta)d(theta)`
`intsec^3thetad(theta)=1/2intsecthetad(theta)+1/2secthetatantheta`
and `intsecthetad(theta)=ln(tantheta+sectheta)`
`intsec^3thetad(theta)=1/2ln(tantheta+sectheta)+1/2secthetatantheta`
`100int(sec^3theta-sectheta)d(theta)=100(1/2secthetatantheta-1/2ln(tantheta+sectheta))`