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

1 Answer
Narad T.
Nov 2, 2016

The answer is =(x+3)/2(sqrt(((x+3)^2-100)))-50ln((sqrt(((x+3)^2)-100)+(x+3))/10)+C=x+32(√((x+3)2−100))−50ln⎛⎜ ⎜ ⎜ ⎜⎝√((x+3)2)−100+(x+3)10⎞⎟ ⎟ ⎟ ⎟⎠+C

Explanation:

Let u=x+3u=x+3 then du=dxdu=dx
int(sqrt((x+3)^2-100))dx=int(sqrt(u^2-100))du∫(√(x+3)2−100)dx=∫(√u2−100)du
Then let u=10secthetau=10secθ=>⇒du=10secthetatanthetadu=10secθtanθ
int(sqrt(u^2-100))du=int(sqrt(100sec^2theta-100))10secthetatantheta(d(theta))∫(√u2−100)du=∫(√100sec2θ−100)10secθtanθ(d(θ))
=100intsecthetatan^2thetad(theta)=100∫secθtan2θd(θ)
=100intsectheta(sec^2theta-1)d(theta)=100∫secθ(sec2θ−1)d(θ)
=100int(sec^3theta-sectheta)d(theta)=100∫(sec3θ−secθ)d(θ)
intsec^3thetad(theta)=1/2intsecthetad(theta)+1/2secthetatantheta∫sec3θd(θ)=12∫secθd(θ)+12secθtanθ
and intsecthetad(theta)=ln(tantheta+sectheta)∫secθd(θ)=ln(tanθ+secθ)
intsec^3thetad(theta)=1/2ln(tantheta+sectheta)+1/2secthetatantheta∫sec3θd(θ)=12ln(tanθ+secθ)+12secθtanθ
100int(sec^3theta-sectheta)d(theta)=100(1/2secthetatantheta-1/2ln(tantheta+sectheta))100∫(sec3θ−secθ)d(θ)=100(12secθtanθ−12ln(tanθ+secθ))

Related questions
See all questions in Integration by Trigonometric Substitution
Impact of this question
1679 views around the world
You can reuse this answer
Creative Commons License