
William W. answered 03/09/24
Top Pre-Calc Tutor
First set up the problem:
Then sub x = (5/2)sec(θ) or sec(θ) = (2/5)x.
Then dx = (5/2)tan(θ)sec(θ) dθ
The integral becomes:
Simplifying:
5∫tan(θ)√(sec2(θ) - 1) dθ
Then applying the Pythagorean Identity tan2(θ) = sec2(θ) - 1 and taking the square root we get:
5∫tan2(θ) dθ
Again, sub sec2(θ) - 1 for tan2(θ) to get:
5∫sec2(θ) - 1 dθ
Take the antiderivative to get:
5tan(θ) - 5θ + C
Using our original trig sub: sec(θ) = (2/5)x re-write as sec(θ) = (2x)/5 and consider the associated triangle where hypotenuse = 2x and adjacent side = 5. That makes the opposite side = √(4x2 - 25).
Using these relationships, tan(θ) = opp/adj = √(4x2 - 25)/5 and it also makes θ = sec-1[(2x)/5]
Sub these in to get the answer you noted.