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

Question

Charles Anctil · Accepted Answer

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

Emma Aldridge · Answer

undefined To integrate ∫sqrt((x+3)^2 - 100) using trigonometric substitution, let:
x + 3 = 10 sec(θ),
dx = 10 sec(θ) tan(θ) dθ,
Then the integral becomes:
∫sqrt(100 sec^2(θ) - 100) * 10 sec(θ) tan(θ) dθ
= 10 ∫tan(θ) sec(θ) * 10 sec(θ) tan(θ) dθ
= 100 ∫tan^2(θ) sec(θ) dθ
= 100 ∫(sec^2(θ) - 1) sec(θ) dθ
= 100 ∫(sec^3(θ) - sec(θ)) dθ
Now integrate term by term:
= 100 (1/3 sec^3(θ) - ln|sec(θ) + tan(θ)|) + C
Now, substitute back for θ:
= 100 (1/3 sec(θ)(sec^2(θ) - 1) - ln|sec(θ) + tan(θ)|) + C
Finally, replace sec(θ) and tan(θ) with their expressions in terms of x:
= 100/3 (x + 3) sqrt((x+3)^2 - 100) - 100 ln| (x + 3)/10 + sqrt((x+3)^2 - 100)/10 | + C