Find ∫1/((x^2+5)^(3/2)) dx
I figured this would be a trig substitution problem, so I set x = sqrt(5)tanθ and dx = sqrt(5)sec^2(θ) dθ
This would lead to:
∫(sqrt(5)sec^2(θ)) / (5tan^2(θ) + 5)^(3/2) dθ
But I'm kinda lost on what to do next. I'm used to breaking up the bottom part if it's a square root, but what would happen if it's to the (3/2) power? I tried typing into calculators and they simplified it like crazy where I don't even understand how they got to the point. Any help is greatly appreciated!
3 answers
x/(5 sqrt(5)) - x^3/(50 sqrt(5)) + (3 x^5)/(1000 sqrt(5)) - x^7/(2000 sqrt(5)) + O(x^9)
atsqrt(5):
(1/2 - i/2)/(5^(3/4) sqrt(x + i sqrt(5))) + ((3/40 + (3 i)/40) sqrt(x + i sqrt(5)))/5^(1/4) + ((1/64 - i/64) (x + i sqrt(5))^(3/2))/5^(3/4) - ((7/6400 + (7 i)/6400) (x + i sqrt(5))^(5/2))/5^(1/4) + O((x + i sqrt(5))^(7/2))
(Puiseux series)
(1/2 - i/2)/(5^(3/4) sqrt(x + i sqrt(5))) + ((3/40 + (3 i)/40) sqrt(x + i sqrt(5)))/5^(1/4) + ((1/64 - i/64) (x + i sqrt(5))^(3/2))/5^(3/4) - ((7/6400 + (7 i)/6400) (x + i sqrt(5))^(5/2))/5^(1/4) + O((x + i sqrt(5))^(7/2))
(Puiseux series)
Woah, I don't think there's a need for i in this problem.
This is how I end up doing the problem:
∫(sqrt(5)sec^2(θ)) / (5tan^2(θ) + 5)^(3/2) dθ
∫(sqrt(5)sec^2(θ)) / (5sqrt(5)sec^3(θ)) dθ
Simplifying..
∫(sec^2(θ)/ (5sec^3(θ)) dθ
(1/5) ∫ (sec^2(θ) / (sec^3(θ)) dθ
(1/5) ∫ 1 / (sec^2(θ)) dθ
Using the property : 1/sec(θ) = cos(θ)
(1/5) ∫ cos(θ) dθ
(1/5)sinθ + C
Integrating back to find x using right triangle trig would lead to the final answer being:
x / (5sqrt(x^2 + 5)) + C
I love math.
This is how I end up doing the problem:
∫(sqrt(5)sec^2(θ)) / (5tan^2(θ) + 5)^(3/2) dθ
∫(sqrt(5)sec^2(θ)) / (5sqrt(5)sec^3(θ)) dθ
Simplifying..
∫(sec^2(θ)/ (5sec^3(θ)) dθ
(1/5) ∫ (sec^2(θ) / (sec^3(θ)) dθ
(1/5) ∫ 1 / (sec^2(θ)) dθ
Using the property : 1/sec(θ) = cos(θ)
(1/5) ∫ cos(θ) dθ
(1/5)sinθ + C
Integrating back to find x using right triangle trig would lead to the final answer being:
x / (5sqrt(x^2 + 5)) + C
I love math.
2 answers